Sequences & Series

Problem 701

e the first term is 9 and the third term is 181\frac{1}{81}. (Why are there two possible answers?) Find the 4 th term in the geometric sequence where the first term is 6 and the 7 th term is 332\frac{3}{32}.
For the geometric sequence 3,m,n,192,3, m, n, 192, \ldots, find the values for mm and nn. Find the value of xx such that the following sequence forms a geometric progression: x1,3x+4,6x+8x-1,3 x+4,6 x+8

See Solution

Problem 702

Use the arithmetic sequence formula to find the 38th number in the sequence: 3,6,93,6,9 \ldots

See Solution

Problem 703

16 / Final (April 25, 2024) - [9 points] The Taylor series centered at x=1x=1 for a function T(x)T(x) is given by: T(x)=n=0(n!)2(5)n(2n)!(x1)4n+3T(x)=\sum_{n=0}^{\infty} \frac{(n!)^{2}}{(-5)^{n} \cdot(2 n)!}(x-1)^{4 n+3} a. [6 points] Find the radius of convergence of the Taylor series above. Show your work. Do not attempt to find the interval of convergence. page 1 x - 9 points] The Taylor series ces T(x)=n=0(n!)2(5)n(2n)!(x1)4n+3 is given by: T(x)=\sum_{n=0}^{\infty} \frac{(n!)^{2}}{(-5)^{n} \cdot(2 n)!}(x-1)^{4 n+3} \text { is given by: } - ((n+1)kε)2(5)2+22(2n+2))(5)1(x - 1)4n+7(25)2(2n)1(x2)α(x1)4n+3(2n+2)(2n+1)(2n)5(2n+1)(n+1)2n!(n+2n+2(n+1))2(n+1)\begin{array}{l} \frac{\left((n+1) \frac{\sqrt{k}}{\varepsilon}\right)^{2}}{(-5)^{\frac{2+2}{2}}} \frac{(2 n+2))}{(-5)} \\ 1^{(x} \\ \text { - } 1)^{4 n+7} \\ \frac{(25)^{2} \cdot(2 n)^{1}}{\left(x^{2}\right)^{\alpha}(x-1)^{4 n+3}} \\ \frac{(2 n+2)(2 n+1)(2 n)}{-5(2 n+1)(n+1)} \\ 2 n! \\ (n+2 n+2 \\ \left.(n+1)^{\prime}\right)^{2(n+1)} \end{array} 26+2=82+2=42+2+6\begin{array}{c} 2 \\ 6+2=8 \\ 2+2=4 \\ 2+2+6 \end{array} 604-60_{4} \cdot 1 b. [3 points] Compute T(123)(1)T^{(123)}(1). Show your work. You do not need to simplify your answer. Answer: T(123)(1)=\quad T^{(123)}(1)= \qquad

See Solution

Problem 704

2) Determine how much of the total loan payment applies toward principal and how much applies toward interest for a student loan of \$38,156 at a fixed APR of 8\% for 11 years. A) \$38,156 pays off the principal and \$19,321.94 represents interest payments. B) \$38,156 pays off the principal and \$19,398.46 represents interest payments. C) \$38,156 pays off the principal and \$19,362.76 represents interest payments. D) \$38,156 pays off the principal and \$19,338.95 represents interest payments.

See Solution

Problem 705

Type the missing number in this sequer 1, 3, 9, 27, 81,

See Solution

Problem 706

Question 3 (1 point) Imagine you need $425\$ 425 to cover unexpected car repairs. You decide to get a payday loan. The payday lender charges a $30\$ 30 fee for a two-week loan. If this loan was to be rolled over for an entire year then how much would be owed?
Your Answer: \square Answer

See Solution

Problem 707

[10] 3. For f(x)=32x+75f(x)=-32 x+75 and g(x)=7(0.6)xg(x)=7(0.6)^{x} find the following by only using a proper formula: (A) k=069f(k)=f(0)+f(1)+f(2)++f(69)\sum_{k=0}^{69} f(k)=f(0)+f(1)+f(2)+\cdots+f(69). (B) h=044g(h)=g(0)+g(1)+g(2)++g(44)\sum_{h=0}^{44} g(h)=g(0)+g(1)+g(2)+\cdots+g(44). (C) What is h=0g(h)=g(0)+g(1)+g(2)++g(n)+g(n+1)+\sum_{h=0}^{\infty} g(h)=g(0)+g(1)+g(2)+\cdots+g(n)+g(n+1)+\ldots ?

See Solution

Problem 708

An investor deposits \$1200 yearly in an annuity at 7% interest. Find the total value after 1, 2, and 3 years.

See Solution

Problem 709

Keiko deposits \950annuallyina6950 annually in a 6% annuity. Find values at the end of years 1, 2, and 3: (a) V_1,(b), (b) V_2,(c), (c) V_3$.

See Solution

Problem 710

Ravi deposits \$625 yearly in an annuity at 8% interest. Find its value at the end of years 1, 2, and 3.

See Solution

Problem 711

Write a function for the arithmetic sequence of Kaia's savings: $525\$ 525, $580\$ 580, $635\$ 635, $690\$ 690.

See Solution

Problem 712

Explain the difference between a sequence and a series. A sequence is a function whose domain is the set of positive integers. A series is a summation of the terms of a sequence. A sequence is a summation whose domain is the set of positive integers. A series is a function of the terms of a sequence. A series is a summation whose domain is the set of positive integers. A sequence is a function of the terms of a series. A series is a function whose domain is the set of positive integers. A sequence is a summation of the terms of a series.

See Solution

Problem 713

Write a formula for the general term (the nth term) of the arithmetic sequence 5,2,1,4. Then use the formula for an to find a20.\text{Write a formula for the general term (the nth term) of the arithmetic sequence } 5, 2, -1, -4. \text{ Then use the formula for } a_n \text{ to find } a_{20}.

See Solution

Problem 714

Доказательство. Мы воспользуемся леммой 2.1.10 в случае необходимости и тогда можем считать, что неравенства выполнены для всех n0n \geq 0. (1) Имеем для каждого n2n \geq 2 xn=x1x2x1x3x2xn1xn2xnxn1x_{n}=x_{1} \cdot \frac{x_{2}}{x_{1}} \cdot \frac{x_{3}}{x_{2}} \cdots \frac{x_{n-1}}{x_{n-2}} \cdot \frac{x_{n}}{x_{n-1}}
по условию x2x1,x3x2,,xnxn1q<1\frac{x_{2}}{x_{1}}, \frac{x_{3}}{x_{2}}, \ldots, \frac{x_{n}}{x_{n-1}} \leq q<1
тогда xn=x1x2x1x3x2xn1xn2xnxn1x1qn1x_{n}=x_{1} \cdot \frac{x_{2}}{x_{1}} \cdot \frac{x_{3}}{x_{2}} \cdots \frac{x_{n-1}}{x_{n-2}} \cdot \frac{x_{n}}{x_{n-1}} \leq x_{1} q^{n-1}
С другой стороны, (см. пример 2.1.4) ряд ( qnq^{n} ) сходится при q<1q<1, а тогда по предложению 2.1.8 ряд ( x1qnx_{1} q^{n} ) тоже сходится. Наконец, по признаку 1 (Следствие 2.2.2) ряд (xn)\left(x_{n}\right) сходится.

See Solution

Problem 715

\begin{problem} Докажите признаки Даламбера для сходимости и расходимости ряда.
1. Если для почти всех n0n \geq 0 xn+1xnq<1,\frac{x_{n+1}}{x_{n}} \leq q < 1, то ряд (xn)\left(x_{n}\right) сходится; если же для почти всех n0n \geq 0 xn+1xn1,\frac{x_{n+1}}{x_{n}} \geq 1, то ряд (xn)\left(x_{n}\right) расходится.
2. Если limnxn+1xn=q,\lim_{n \rightarrow \infty} \frac{x_{n+1}}{x_{n}} = q, то ряд (xn)\left(x_{n}\right) при q<1q < 1 сходится, а при 1<q1 < q \leq \infty расходится.
\textbf{Доказательство:} Мы воспользуемся леммой 2.1.10 в случае необходимости и тогда можем считать, что неравенства выполнены для всех n0n \geq 0.
(1) Имеем для каждого n2n \geq 2 xn=x1x2x1x3x2xn1xn2xnxn1x_{n} = x_{1} \cdot \frac{x_{2}}{x_{1}} \cdot \frac{x_{3}}{x_{2}} \cdots \frac{x_{n-1}}{x_{n-2}} \cdot \frac{x_{n}}{x_{n-1}} по условию x2x1,x3x2,,xnxn1q<1,\frac{x_{2}}{x_{1}}, \frac{x_{3}}{x_{2}}, \ldots, \frac{x_{n}}{x_{n-1}} \leq q < 1, тогда xn=x1x2x1x3x2xn1xn2xnxn1x1qn1.x_{n} = x_{1} \cdot \frac{x_{2}}{x_{1}} \cdot \frac{x_{3}}{x_{2}} \ldots \frac{x_{n-1}}{x_{n-2}} \cdot \frac{x_{n}}{x_{n-1}} \leq x_{1} q^{n-1}.
С другой стороны, (см. пример 2.1.4) ряд (qn)(q^{n}) сходится при q<1q < 1, а тогда по предложению 2.1.8 ряд (x1qn)(x_{1} q^{n}) тоже сходится. Наконец, по признаку 1 (Следствие 2.2.2) ряд (xn)(x_{n}) сходится.
(2) Если же x2x1,x3x2,,xnxn1>1,\frac{x_{2}}{x_{1}}, \frac{x_{3}}{x_{2}}, \ldots, \frac{x_{n}}{x_{n-1}} > 1, то xnx1,n2.x_{n} \geq x_{1}, \quad n \geq 2.
Ряд (yn)\left(y_{n}\right), где все yn=x1y_{n} = x_{1}, очевидно, расходится, тогда по признаку 1 (Следствие 2.2.2), ряд (xn)\left(x_{n}\right) тоже расходится.
(3) Пусть limnxn+1xn=q\lim_{n \rightarrow \infty} \frac{x_{n+1}}{x_{n}} = q, тогда по определению 1.2.5 для любого ε>0\varepsilon > 0 существует такой NN, что для всех nNn \geq N имеют место неравенства qε<xn+1xn<q+ε.q - \varepsilon < \frac{x_{n+1}}{x_{n}} < q + \varepsilon.
Если q<1q < 1, то пусть q+ε<1q + \varepsilon < 1, тогда для всех nN,xn+1xn1n \geq N, \frac{x_{n+1}}{x_{n}} \leq 1, тогда по доказанному признаку (1) ряд (xn)(x_{n}) сходится.
Если q>1q > 1, то возьмём ε>0\varepsilon > 0 такое, что qε>1q - \varepsilon > 1. Но xn+1xn>qε\frac{x_{n+1}}{x_{n}} > q - \varepsilon при каких-то nNn \geq N, поэтому для Nn0<nN \leq n_{0} < n получаем xn=xnxn1xn1xn2xn0+1xn0xn0>(qε)nn0xn0.x_{n} = \frac{x_{n}}{x_{n-1}} \cdot \frac{x_{n-1}}{x_{n-2}} \cdots \frac{x_{n_{0}+1}}{x_{n_{0}}} x_{n_{0}} > (q - \varepsilon)^{n-n_{0}} x_{n_{0}}. \end{problem}

See Solution

Problem 716

2.2.3 Инвариантность суммы
Докажем инвариантность суммы сходящегося положительного ряда при произвольной перестановки его элементов.
Теорема 2.2.10. Пусть (xn)\left(x_{n}\right) - сходящийся положительный ряд с суммой S. Тогда полученный в результате произвольной перестановки его элементов новый (заново перенумерованный) ряд также сходится и имеет ту же сумму s. 38
Доказательство. Пусть x1=xn1,,xk=xnk,x_{1}^{\prime}=x_{n_{1}}, \ldots, x_{k}^{\prime}=x_{n_{k}}, \ldots, и пусть n:=max{n1,,nk}n:=\max \left\{n_{1}, \ldots, n_{k}\right\}, рассмотрим тогда частичные суммы Sn:=x1+x+xn, Sk:=x1++xk,\mathrm{S}_{n}:=x_{1}+x \cdots+x_{n}, \quad \mathrm{~S}_{k}^{\prime}:=x_{1}^{\prime}+\cdots+x_{k}^{\prime},
так как 1n1,,nkn1 \leq n_{1}, \ldots, n_{k} \leq n и (xn)\left(x_{n}\right) - положительный ряд, то SkSnS_{k}^{\prime} \leq S_{n}
Но, положительный ряд ( xnx_{n} ) сходится, а тогда по критерию сходимости положительного ряда (см. Теорема 2.2.1), последовательность (Sn)\left(\mathrm{S}_{n}\right) ограничена, и более того SnS\mathrm{S}_{n} \leq \mathrm{S} для всех nn. Таким образом, для всех kk получаем SkSnS,\mathrm{S}_{k}^{\prime} \leq \mathrm{S}_{n} \leq \mathrm{S}, m.e., последовательность (Sk)\left(S_{k}^{\prime}\right) частичных сумм ряда (xk)\left(x_{k}^{\prime}\right) ограничена, а тогда по критерию

See Solution

Problem 717

Доказательство. Воспользовавшись леммой 2.1.10, мы можем считать, что anan+1\left|a_{n}\right| \geq\left|a_{n+1}\right| для всех nn. Для удобства положим, что первый элемент ряда - это a0a_{0}, m.e. n0n \geq 0. Рассмотрим частичную сумму S2n+1S_{2 n+1}, имеем S2n+1=a0a1+a2a3+a4++a2na2n+1=a0(a1a2)(a3a4)(a2n1a2n)a2n+1\begin{aligned} \mathrm{S}_{2 n+1} & =\left|a_{0}\right|-\left|a_{1}\right|+\left|a_{2}\right|-\left|a_{3}\right|+\left|a_{4}\right|+\cdots+\left|a_{2 n}\right|-\left|a_{2 n+1}\right| \\ & =\left|a_{0}\right|-\left(\left|a_{1}\right|-\left|a_{2}\right|\right)-\left(\left|a_{3}\right|-\left|a_{4}\right|\right)-\cdots-\left(\left|a_{2 n-1}\right|-\left|a_{2 n}\right|\right)-\left|a_{2 n+1}\right| \end{aligned}
так как anan+1\left|a_{n}\right| \geq\left|a_{n+1}\right|, то каждая скобка положительна, это значит, что S2n+1a0S_{2 n+1} \leq\left|a_{0}\right|, т.е. последовательность ( S2n+1\mathrm{S}_{2 n+1} ) ограничена сверху.
С другой стороны, мы можем записать S2n+1=a0a1+a2a3++a2n2a2n1+a2na2n+1=(a0a1)+(a2a3)++(a2n2a2n1)+(a2na2n+1)=S2n1+(a2na2n+1)\begin{aligned} \mathrm{S}_{2 n+1} & =\left|a_{0}\right|-\left|a_{1}\right|+\left|a_{2}\right|-\left|a_{3}\right|+\cdots+\left|a_{2 n-2}\right|-\left|a_{2 n-1}\right|+\left|a_{2 n}\right|-\left|a_{2 n+1}\right| \\ & =\left(\left|a_{0}\right|-\left|a_{1}\right|\right)+\left(\left|a_{2}\right|-\left|a_{3}\right|\right)+\cdots+\left(\left|a_{2 n-2}\right|-\left|a_{2 n-1}\right|\right)+\left(\left|a_{2 n}\right|-\left|a_{2 n+1}\right|\right) \\ & =S_{2 n-1}+\left(\left|a_{2 n}\right|-\left|a_{2 n+1}\right|\right) \end{aligned}
и так как a2na2n+1\left|a_{2 n}\right| \geq\left|a_{2 n+1}\right|, то S2n+1S2n1S_{2 n+1} \geq S_{2 n-1}, т.е. она не убывает. Итак, последовательность (S2n+1)\left(S_{2 n+1}\right) ограничена сверху и не убывает, тогда по теореме Вейерштрасса 1.4 .7 у неё есть предел limn S2n+1=Sa0\lim _{n \rightarrow \infty} \mathrm{~S}_{2 n+1}=\mathrm{S} \leq\left|a_{0}\right|.
Наконец, мы также можем записать S2n+1=a0a1+a2a3+a2na2n+1=S2na2n+1\begin{aligned} S_{2 n+1} & =\left|a_{0}\right|-\left|a_{1}\right|+\left|a_{2}\right|-\left|a_{3}\right|+\left|a_{2 n}\right|-\left|a_{2 n+1}\right| \\ & =S_{2 n}-\left|a_{2 n+1}\right| \end{aligned}
так как limn S2n+1=S\lim _{n \rightarrow \infty} \mathrm{~S}_{2 n+1}=\mathrm{S} и по условию limna2n+1=0\lim _{n \rightarrow \infty}\left|a_{2 n+1}\right|=0, то по теореме 1.3.1 limnS2n=limn(S2n+1+a2n+1)=S+0=S\lim _{n \rightarrow \infty} S_{2 n}=\lim _{n \rightarrow \infty}\left(S_{2 n+1}+\left|a_{2 n+1}\right|\right)=S+0=S
Итак, мы показали, что limn Sn=S\lim _{n \rightarrow \infty} \mathrm{~S}_{n}=\mathrm{S}, что и означает сходимость ряда.

See Solution

Problem 718

Доказательство. Воспользовавшись леммой 2.1.10, мы можем считать, что anan+1\left|a_{n}\right| \geq\left|a_{n+1}\right| для всех nn. Для удобства положим, что первый элемент ряда - это a0a_{0}, m.e. n0n \geq 0. Рассмотрим частичную сумму S2n+1S_{2 n+1}, имеем S2n+1=a0a1+a2a3+a4++a2na2n+1=a0(a1a2)(a3a4)(a2n1a2n)a2n+1\begin{aligned} \mathrm{S}_{2 n+1} & =\left|a_{0}\right|-\left|a_{1}\right|+\left|a_{2}\right|-\left|a_{3}\right|+\left|a_{4}\right|+\cdots+\left|a_{2 n}\right|-\left|a_{2 n+1}\right| \\ & =\left|a_{0}\right|-\left(\left|a_{1}\right|-\left|a_{2}\right|\right)-\left(\left|a_{3}\right|-\left|a_{4}\right|\right)-\cdots-\left(\left|a_{2 n-1}\right|-\left|a_{2 n}\right|\right)-\left|a_{2 n+1}\right| \end{aligned}
так как anan+1\left|a_{n}\right| \geq\left|a_{n+1}\right|, то каждая скобка положительна, это значит, что S2n+1a0S_{2 n+1} \leq\left|a_{0}\right|, т.е. последовательность ( S2n+1\mathrm{S}_{2 n+1} ) ограничена сверху.
С другой стороны, мы можем записать S2n+1=a0a1+a2a3++a2n2a2n1+a2na2n+1=(a0a1)+(a2a3)++(a2n2a2n1)+(a2na2n+1)=S2n1+(a2na2n+1)\begin{aligned} \mathrm{S}_{2 n+1} & =\left|a_{0}\right|-\left|a_{1}\right|+\left|a_{2}\right|-\left|a_{3}\right|+\cdots+\left|a_{2 n-2}\right|-\left|a_{2 n-1}\right|+\left|a_{2 n}\right|-\left|a_{2 n+1}\right| \\ & =\left(\left|a_{0}\right|-\left|a_{1}\right|\right)+\left(\left|a_{2}\right|-\left|a_{3}\right|\right)+\cdots+\left(\left|a_{2 n-2}\right|-\left|a_{2 n-1}\right|\right)+\left(\left|a_{2 n}\right|-\left|a_{2 n+1}\right|\right) \\ & =S_{2 n-1}+\left(\left|a_{2 n}\right|-\left|a_{2 n+1}\right|\right) \end{aligned}
и так как a2na2n+1\left|a_{2 n}\right| \geq\left|a_{2 n+1}\right|, то S2n+1S2n1S_{2 n+1} \geq S_{2 n-1}, т.е. она не убывает. Итак, последовательность (S2n+1)\left(S_{2 n+1}\right) ограничена сверху и не убывает, тогда по теореме Вейерштрасса 1.4 .7 у неё есть предел limn S2n+1=Sa0\lim _{n \rightarrow \infty} \mathrm{~S}_{2 n+1}=\mathrm{S} \leq\left|a_{0}\right|.
Наконец, мы также можем записать S2n+1=a0a1+a2a3+a2na2n+1=S2na2n+1\begin{aligned} S_{2 n+1} & =\left|a_{0}\right|-\left|a_{1}\right|+\left|a_{2}\right|-\left|a_{3}\right|+\left|a_{2 n}\right|-\left|a_{2 n+1}\right| \\ & =S_{2 n}-\left|a_{2 n+1}\right| \end{aligned}
так как limn S2n+1=S\lim _{n \rightarrow \infty} \mathrm{~S}_{2 n+1}=\mathrm{S} и по условию limna2n+1=0\lim _{n \rightarrow \infty}\left|a_{2 n+1}\right|=0, то по теореме 1.3.1 limnS2n=limn(S2n+1+a2n+1)=S+0=S\lim _{n \rightarrow \infty} S_{2 n}=\lim _{n \rightarrow \infty}\left(S_{2 n+1}+\left|a_{2 n+1}\right|\right)=S+0=S
Итак, мы показали, что limn Sn=S\lim _{n \rightarrow \infty} \mathrm{~S}_{n}=\mathrm{S}, что и означает сходимость ряда.

See Solution

Problem 719

Предложение 2.3.10. Если ряд абсолютно сходится, то при любой перестановке его элементов абсолютная сходимость полученного нового ряда не нарушается и более того, его сумма остаётся прежней.
Доказательство. Пусть ряд (xn)\left(x_{n}\right) сходится абсолютно, рассмотрим ряды (xn+),(xn)\left(x_{n}^{+}\right),\left(x_{n}^{-}\right)(конструкция 2.3.8), очевидно, что xn=xn+xnx_{n}=x_{n}^{+}-x_{n}^{-}для всех nn. Так как ряд ( xnx_{n} ) сходится абсолютно, то ввиду xn+xn,xnxnx_{n}^{+} \leq\left|x_{n}\right|, x_{n}^{-} \leq\left|x_{n}^{-}\right|и признака сравнения (теорема 2.2.2) ряды (xn+),(xn)\left(x_{n}^{+}\right),\left(x_{n}^{-}\right) тоже сходятся.
Пусть ряд, полученный после перестановки исходного ряда, имеет вид ( yny_{n} ), рассмотрим также ряды (yn+),(yn)\left(y_{n}^{+}\right),\left(y_{n}^{-}\right)(Конструкция 2.3.8), тогда yn=yn+yny_{n}=y_{n}^{+}-y_{n}^{-}, и мы получаем S=S+S=n=1xn+n=1xn=n=1yn+n=1yn=n=1(yn+yn)=n=1yn.\begin{aligned} S & =S^{+}-S^{-} \\ & =\sum_{n=1}^{\infty} x_{n}^{+}-\sum_{n=1}^{\infty} x_{n}^{-} \\ & =\sum_{n=1}^{\infty} y_{n}^{+}-\sum_{n=1}^{\infty} y_{n}^{-} \\ & =\sum_{n=1}^{\infty}\left(y_{n}^{+}-y_{n}^{-}\right) \\ & =\sum_{n=1}^{\infty} y_{n} . \end{aligned} (по предложению 2.3.2) (по теореме 2.2.10) (по предложению 2.1.8)
Что и требовалось доказать.

See Solution

Problem 720

DEPARTMENT: SCIENCE LABORATORY TITLE: MTH 101
The nthn^{th} term of the sequence is given as 322n3 \cdot 2^{2n}. Obtain the first four term of Sequence.

See Solution

Problem 721

A finite sequence is shown. {25,22,19,,32}\{-25,-22,-19, \ldots, 32\}
Which sigma notation can be used to represent the series for the finite sequence? Help: Introduction to Sigma Notation (video). n=118(3n28)\sum_{n=1}^{18}(3 n-28) n=120(3n28)\sum_{n=1}^{20}(3 n-28) n=120(3n22)\sum_{n=1}^{20}(-3 n-22) n=118(3n22)\sum_{n=1}^{18}(-3 n-22)

See Solution

Problem 722

```latex \text{Part I-Write the most simplified form of your answer in the space provided (3 points)}
\text{a) Determine whether the following sequence converges or diverges.}
\text{a. } \sum_{n=0}^{\infty} \frac{2^{2}}{(x+1)^{2}} \text{ is } \qquad
\text{b. } \sum_{n=1}^{\infty} \frac{3}{n^{2}-3n+2} \text{ is } \qquad
\qquad \text{convergent, or divergent} \qquad
\text{1. For what values of } x \text{ does the series } \sum_{n=0}^{\infty} n x^{n} \text{ converge?} \qquad
\text{Part II: Work out each of the following clearly and neatly showing all the necessary steps: (4 points each)}
\text{5. Determine if the following series converges or diverges.}
\text{a. } \sum_{i=3}^{\infty}\left(1-\frac{3}{n}\right) m^{2} \text{ by root test.}
\text{b. } \sum^{\infty}=\frac{n}{\sqrt{\pi^{2}-12}} \text{ by comparison test.} ```

See Solution

Problem 723

เi. on 5 - Notes C. Q Line Guide Reset Answer
This question has multiple parts. Be sure to answer all the parts of this question.
Each figure is created using green hexagon tiles. PART A
Is the sequence describing the number of green hexagons used in each figure an arithmetic or geometric sequence? Explain.
Figure 1 Figure 2 Figure 3 Enter your response here

See Solution

Problem 724

Determine the monthly payment for the installment loan. \begin{tabular}{|l|l|l|l|} \hline Amount & \begin{tabular}{l} Annual \\ Percentage \end{tabular} & \begin{tabular}{l} Number of \\ Payments per \\ Year (n)(\mathbf{n}) \end{tabular} & \begin{tabular}{l} Time in \\ Years (t)(t) \end{tabular} \\ \hline Financed (P) & Rate (r)(\boldsymbol{r}) & 12 & 3 \\ \hline$14,000\$ 14,000 & 4%4 \% & & \\ \hline \end{tabular}
Click the icon to view the partial APR table.
The monthly payment is \ \square$ (Round to the nearest cent as needed.)

See Solution

Problem 725

Determine the monthly payment for the installment loan. \begin{tabular}{|c|c|c|c|} \hline \begin{tabular}{l} Amount \\ Financed (P) \end{tabular} & Annual Percentage Rate (r) & Number of Payments per Year (n) & Time in Years (t) \\ \hline \$14,000 & 4\% & 12 & 3 \\ \hline \end{tabular}
Click the icon to view the partial APR table.
The monthly payment is $\$ \square (Round to the nearest cent as needed.) Finance Rates

See Solution

Problem 726

Question 3 4 Pc
Use the Principle of Mathematical Induction to select the induction step for the proof of the logical statement. P(n):2+7+12+.+(5n3)=n2(5n1)P(n): 2+7+12+\ldots .+(5 n-3)=\frac{n}{2}(5 n-1), for nNn \in N. Therefore, for some kNk \in N, the induction step would be given by (A) k2(5k1)+(5k+2)=k+12(5(k+1)1)\frac{k}{2}(5 k-1)+(5 k+2)=\frac{k+1}{2}(5(k+1)-1) (B) k2(5k1)+k=k2(5k3)\frac{k}{2}(5 k-1)+k=\frac{k}{2}(5 k-3) (C) (5k3)+(k+1)=2(3k1)(5 \mathrm{k}-3)+(\mathrm{k}+1)=2(3 \mathrm{k}-1) (D) k(5k1)+k=5k2k(5 k-1)+k=5 k^{2}

See Solution

Problem 727

3.12. f(x)=3x,2<x<2,l=2f(x)=3-x,-2<x<2, l=2. (Oтвет: 3x=3-x= =2+4πn=1((1)nnsinnπx2).)\left.=2+\frac{4}{\pi} \sum_{n=1}^{\infty}\left(\frac{(-1)^{n}}{n} \sin \frac{n \pi x}{2}\right) .\right)

See Solution

Problem 728

Ray Flagg took out a 60 -month fixed installment loan of $12,000\$ 12,000 to open a new pet store. He paid no money down and began making monthly payments of $232\$ 232. Ray's business does better than expected and instead of making his 36 th payment, Ray wishes to repay his loan in full. Complete parts a) through c).
Click the icon to view the table of interest rates. a) Determine the APR of the installment loan. APR=6.0%\mathrm{APR}=6.0 \% b) How much interest will Ray save by paying off the loan early? (Use the actuarial method)
Interest saved =$333.44=\$ 333.44 (Round to the nearest cent.) c) What is the total amount due to pay off the loan?
The total amount due =$=\$ \square (Round to the nearest cent.) table of interest rates \begin{tabular}{|ccccccccccc|} \hline & \multicolumn{8}{c|}{ Annual Percentage Rate } \\ \cline { 2 - 10 } Number of & 3.0%3.0 \% & 3.5%3.5 \% & 4.0%4.0 \% & 4.5%4.5 \% & 5.0%5.0 \% & 5.5%5.5 \% & 6.0%6.0 \% & 6.5%6.5 \% & 7.0%7.0 \% \\ \hline \multicolumn{7}{c|}{ (Finance charge per $100\$ 100} & of amount financed) \\ Payments & 3.15 & 3.69 & 4.22 & 4.75 & 5.29 & 5.83 & 6.37 & 6.91 & 7.45 \\ 24 & 3.92 & 4.58 & 5.25 & 5.92 & 6.59 & 7.26 & 7.94 & 8.61 & 9.30 \\ 30 & 4.69 & 5.49 & 6.29 & 7.09 & 7.90 & 8.71 & 9.52 & 10.34 & 11.16 \\ 36 & 6.24 & 7.31 & 8.38 & 9.46 & 10.54 & 11.63 & 12.73 & 13.83 & 14.94 \\ 48 & 7.81 & 9.15 & 10.50 & 11.86 & 13.23 & 14.61 & 16.00 & 17.40 & 18.81 \\ 60 & & & & & & & & \end{tabular}

See Solution

Problem 729

اذا كان f(z)=ezezz4f(z)=\frac{e^{z}-e^{-z}}{z^{4}} أوجد مايلي:
1. متسلسلة لورانت للاقتران (z) f(z
2. اوجد قيم z بحيث تكون متسلسلة لورانت متقاربة.

See Solution

Problem 730

اذا كان f(z)=ezezz4f(z)=\frac{e^{z}-e^{-z}}{z^{4}} أوجد مايلي:
1. متسلسلة لورانت للاقتبان (z)
2. اوجد قيم z بحيث تكون متسلسلة لورانت متقاربة.

See Solution

Problem 731

Which represents the explicit formula for the arithmetic sequence an=15+5(n1)a_{n}=15+5(n-1) in function form? f(n)=5n+10f(n)=5 n+10 f(n)=n+10f(n)=n+10 f(n)=n+20f(n)=n+20 f(n)=5n+15f(n)=5 n+15 Previous

See Solution

Problem 732

Prove PnP_{n}: k=1n1k(k+1)=nn+1\sum_{k=1}^{n} \frac{1}{k(k+1)}=\frac{n}{n+1} for all positive integers nn. What’s the first induction step?

See Solution

Problem 733

\text{Show that the series } \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}} \text{ converges.}

See Solution

Problem 734

3 8 points
In units of hRh R, what is the amount of energy associated with the transition from n=4n=4 to n=1n=1 in the hydrogen emission spectrum? ( h is Planck's constant and R=3.3×1015 HzR=3.3 \times 10^{15} \mathrm{~Hz} in the Rydberg equation, although you don't need to use these numbers in this problem.) 116hR\frac{1}{16} h R 34hR\frac{3}{4} h R 14hR\frac{1}{4} h R 1516hR\frac{15}{16} h R 4 8 points

See Solution

Problem 735

Considérons la fonction numérique ff de la variable réelle xx définie sur R\mathbb{R} par: f(x)=12x2+3f(x)=\frac{1}{2} \sqrt{x^{2}+3}
1. a) Montrer que: f(x)=x2x2+3f^{\prime}(x)=\frac{x}{2 \sqrt{x^{2}+3}} pour tout x0x \geq 0, puis dresser le tableau de variations de la fonction ff sur R\mathbb{R}. a) Montrer que : f([0;1][0;1[f([0 ; 1] \subset[0 ; 1[.
2. a) Résoudre dans R\mathbb{R} l'équation (E):f(x)=x(E): f(x)=x. b) Montrer que : f(x)>xf(x)>x pour tout élément xx de l'intervalle [0;1[[0 ; 1[.
3. On considère la suite (Un)\left(U_{n}\right) définie par : U0=0U_{0}=0 et Un+1=f(Un)U_{n+1}=f\left(U_{n}\right) pour tout entier naturel nn. 3.1. Montrer par récurrence que : 0Un<10 \leq U_{n}<1 pour tout entier naturel nn. 3.2. Etudier le sens de variations de la suite (Un)\left(U_{n}\right). 3.3. Montrer que la suite (Un)\left(U_{n}\right) est convergente et déterminer sa limite.

See Solution

Problem 737

( 20 علامة) r=1n2r1=n2\sum_{r=1}^{n} 2 r-1=n^{2}
السؤال الثالث:
برهن باستخدام الاستقراء الرياضي أن:

See Solution

Problem 738

30.000+t=168.000(1+0,12)t-30.000+\sum_{t=1}^{6} \frac{8.000}{(1+0,12)^{t}}

See Solution

Problem 739

1. Find F30F_{30}.
2. Calculate F10+F20+F5F_{10} + F_{20} + F_{5}.
3. Solve 3F182F103 F_{18} - 2 F_{10}.
4. Determine F215F7F_{21} - 5 F_{7}.
5. Find 4F5+F112\frac{4 F_{5} + F_{11}}{2}.
6. Calculate 3F136\frac{3 F_{13}}{6}.
7. What is F750\mathrm{F}_{75} * 0?
8. Compute 10 F6310 \mathrm{~F}_{6} * 3.
9. Find F352\frac{F_{35}}{2}.
10. Calculate F2933\frac{F_{29}}{3} * 3.

See Solution

Problem 740

Find the explicit formula for the arithmetic sequence starting with a1=26a_{1}=26 and the terms 26,33,40,47,54,61,26,33,40,47,54,61,\ldots.

See Solution

Problem 741

Question 1 (a) Kapil opened a recurring deposit account in a bank. He deposits ₹ 1500 every month [3] for 2 years at 5%5 \% simple interest per annum. Find the total interest earned by Kapil on maturity. b) If A=[2112],B=[1423]A=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right] and C=[1225]C=\left[\begin{array}{ll}-1 & 2 \\ -2 & 5\end{array}\right], find A(BC)A(B-C). [3]
The table below shows the daily expenditure on food of 50 house-holds in a locality. [4] \begin{tabular}{|c|c|c|c|c|c|c|} \hline \begin{tabular}{c} Daily \\ Expenditure \\ (in ₹) \end{tabular} & 01000-100 & 100200100-200 & 200300200-300 & 300400300-400 & 400500400-500 & 500600500-600 \\ \hline \begin{tabular}{c} Number of \\ House-holds \end{tabular} & 5 & 8 & 15 & 10 & 7 & 5 \\ \hline \end{tabular}
Using graph paper, draw a histogram representing the above distribution and estimate the mode. Take along xx-axis 2 cm=1002 \mathrm{~cm}=₹ 100 and along yy-axis 2 cm=22 \mathrm{~cm}=2 Households.
This paper consists of 8 printed pages. 11 Turn Ov yright reserved.

See Solution

Problem 742

1. Is the series n=1n10001.001n\sum_{n=1}^{\infty} \frac{n^{1000}}{1.001^{n}} convergent or divergent?

See Solution

Problem 743

Given two terms in an arithmetic sequence find the recursive formula. 27) a18=3362a_{18}=3362 and a38=7362a_{38}=7362 28) a18=44.3a_{18}=44.3 and a33=84.8a_{33}=84.8

See Solution

Problem 744

Find the arithmetic sequence for an=5n+6a_{n}=5n+6 starting with n=1n=1.

See Solution

Problem 745

Find the arithmetic sequence for n=19n=19 given the formula an=6n+0a_{n}=6n+0.

See Solution

Problem 746

Prove by induction that for positive integers nn:
1+2(12)+3(12)2++n(12)n1=4n+22n11 + 2(\frac{1}{2}) + 3(\frac{1}{2})^2 + \ldots + n(\frac{1}{2})^{n-1} = 4 - \frac{n+2}{2^{n-1}}.

See Solution

Problem 747

Marissa deposits \$1,000 and adds \$100 monthly for 40 years at a 6.5% annual return. What's her total balance?

See Solution

Problem 748

Find fixed points of the sequence defined by an+1=12(an+3an)a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{3}{a_{n}}\right).

See Solution

Problem 749

Find the 11th 11^{\text {th }} term of the sequence defined by 2n232 n^{2}-3.

See Solution

Problem 750

Find the first negative term tt and an expression for the nthn^{\text{th}} term of the sequence: 32, 26, 20, 14, 8.

See Solution

Problem 751

Write an equation to describe the sequence below. Use nn to represent the position of a term in the sequence, where n=1n=1 for the first term. 34,102,306,34,-102,306, \ldots
Write your answer using decimals and integers. an=()n1a_{n}=\square(\square)^{n-1} Submit

See Solution

Problem 752

Find the right number that fits the sequence. 7410121336?\begin{array}{lllllll}7 & 4 & 10 & 12 & 13 & 36 & ?\end{array}
49 25 16 15 72 Continue

See Solution

Problem 753

Soit le sinte μn\mu_{n} sefin por {μ0=4μn+1=12μn+5\left\{\begin{array}{l}\mu_{0}=4 \\ \mu_{n+1}=\frac{1}{2}\end{array} \mu_{n}+5\right. a) calculer μ1,μ2,μ3\mu_{1}, \mu_{2}, \mu_{3}. b) justifier que (un) ni erthimethique ni Gcométrique o) expore vn=μn10v_{n}=\mu_{n}-10. a) Montre que (vn)(SG)\left(v_{n}\right)(S \cdot G). b) Exprimer vnv_{n} puis unu_{n} enfonction den d) oxpose Sn=k=0n1vkS_{n}=\sum_{k=0}^{n-1} v_{k} et Sn=k=0n1μkS_{n}^{\prime}=\sum_{k=0}^{n-1} \mu_{k} a) experime SnS_{n} pins sn'enfonction dek. b) aluiler S325S_{325} et S2024S_{2024}.

See Solution

Problem 754

Write an expression to describe the sequence below. Use nn to represent the position of a term Questions answered in the sequence, where n=1n=1 for the first term. 65,64,63,62,-65,-64,-63,-62, \ldots 13

See Solution

Problem 755

Select the correct answer from each drop-down menu.
When she was 20, Liz started saving $6,000\$ 6,000 a year for retirement. Her goal is to reach $100,000\$ 100,000 in savings by the time she's 30 . Her account earns 8%8 \% interest per year, compounded annually. Liz \square have saved $100,000\$ 100,000 by age 30 . She'll \square her goal by about \square Reset Next

See Solution

Problem 756

Determine if the sequence 200,40,8,200, 40, 8, \ldots is arithmetic, geometric, or neither.

See Solution

Problem 757

Determine if the sequence defined by f(1)=10,f(n)=f(n1)1.5f(1)=10, f(n)=f(n-1)-1.5 for n2n \geq 2 is arithmetic, geometric, or neither.

See Solution

Problem 758

1. Studiare la monotonia e determinare estremo inferiore e superiore della successione an=cos(nπ)+nn2+arctan(n) per nN\{0}a_{n}=\frac{\cos (n \pi)+n}{n^{2}+\arctan (n)} \quad \text { per } n \in \mathbb{N} \backslash\{0\} specificando se sono minimo e massimo.

See Solution

Problem 759

Quiz, December 17, 2024 B1 Is the series n=1ln(1+1n)(n+5n)\sum_{n=1}^{\infty} \ln \left(1+\frac{1}{n}\right) \cdot(\sqrt{n+5}-\sqrt{n}) convergent?

See Solution

Problem 760

20. What is the fourth term in the expansion of (a+b)4(a+b)^{4} ? a. 4a3b4 a^{3} b b. a4a^{4} c. 6a2b26 a^{2} b^{2} d. 4ab34 a b^{3}

See Solution

Problem 761

Question Write the following using sigma notation and then evaluate. The sum of the terms 2k1+42^{k-1}+4 for k=2,3,5k=2,3, \ldots 5.
Provide your answer below:
\square \square k=k= \square

See Solution

Problem 762

4) L1{1s1+1s+1}=L^{-1}\left\{\frac{1}{s-1}+\frac{1}{s+1}\right\}= a) 2sin(t)2 \sin (t) b) 2cos(t)2 \cos (t) c) 2sinh(t)2 \sinh (t) d) 2cosh(t)2 \cosh (t)

See Solution

Problem 763

The terms of the increasing arithmetic sequence ana_{n} are positive. The terms of the increasing geometric sequence gng_{n} are positive. The values of the first terms of both sequences are the same, and the values of the fourth terms of both sequences are the same. Which of the following statements describes the values of the second terms of the sequences?
A The second term of the arithmetic sequence must be less than the second term of the geometric sequence.
B The second term of the arithmetic sequence must be greater than the second term of the geometric sequence. (C) The second term of the arithmetic sequence must be the same value as the second term of the geometric sequence. (D) The relationship between the values of the second terms cannot be determined from the given information.

See Solution

Problem 764

Describe how to generate the next terms in the pattern: A: 5,8,11,14,17,5, 8, 11, 14, 17, \ldots

See Solution

Problem 765

Identify the rule for generating terms in these sequences: a. Pattern A: 5, 8, 11, 14, 17, ... b. Pattern B: 12,1,2,4,8,...\frac{1}{2}, 1, 2, 4, 8, ...

See Solution

Problem 766

6. Assume a business is deciding whether to invest in a new project that is projected to generate profits of $90,000\$ 90,000 each year for the next three years. The project start-up costs are $225,000\$ 225,000. (A) If the business normally earns 11 percent on its investments, should the business invest? Show/explain. (B) If the business normally earns 5 percent on its investments, should the business invest? Show/explain.

See Solution

Problem 767

e) n=17n4n3\sum_{n=1}^{\infty} \frac{7^{n}}{4^{n}-3}

See Solution

Problem 768

2024-12
1. k=1k+32k+1\sum_{k=1}^{\infty} \frac{\sqrt{k}+3}{2 \sqrt{k}+1} 2k=1k2+k7k+12 \sum_{k=1}^{\infty} \frac{k^{2}+k}{7 k+1}
3. k=15k+44k+3\sum_{k=1}^{\infty} \frac{5 k+4}{4 k+3}
4. k=1k+4k2+5\sum_{k=1}^{\infty} \frac{k+4}{k^{2}+5}
5. k=11k+7\sum_{k=1}^{\infty} \frac{1}{\sqrt{k+7}} a. 3 and 5 b. 1 and 3 c. 2 and 4

See Solution

Problem 769

A formula for the sequence an:2,14,98,686,a_{n}: 2,14,98,686, \cdots, where n0n \geq 0 is
Select one: 72n7 \cdot 2^{n} 2+7n2+7 n 27n2 \cdot 7^{n} 7+2n7+2^{n}

See Solution

Problem 770

12. If f(x)=2cosxf(x)=2 \cos x. Find T6(x,π)T_{6}(x, \pi).

See Solution

Problem 771

The input voltage X(t)\mathrm{X}_{(t)} ), and output voltage Y(t)\mathrm{Y}_{(t)} of an electrical system are sampled simultaneously at regular intervals with the following results. n=0,1,2,3,4,5,6,7,8,9,X(nT)=15,10,6,2,1,0,0,0,0,0,Y(nT)=15,15,7.5,2.75,2.5,,,,\begin{array}{r} n=0,1,2, \quad 3, \quad 4,5,6,7,8,9, \ldots \\ X(n T)=15,10,6,2, \quad 1,0,0,0,0,0, \ldots \\ Y(n T)=15,15,7.5,-2.75,-2.5, \cdots, \cdots,-,-\ldots \end{array}
Calculate the missing values of the output voltage Y(nT)\mathrm{Y}(\mathrm{nT}) above. Mint: Assume that X(t)=0\mathrm{X}(\mathrm{t})=0 for t<0\mathrm{t}<0, and that only nonzero imputse response samples are h(n)\mathrm{h}(\mathrm{n}), for 0n40 \leq \mathrm{n} \leq 4.

See Solution

Problem 772

8.1 .00 9.1 .00 10.1 .00 11.1 .00 12.1 .00 13.1 .00 14.1 .00

See Solution

Problem 773

Exercice 4 Soit la suite (Un)\left(U_{n}\right) définiepar {U0=23Un+1=12Un+n22+12\left\{\begin{array}{c}U_{0}=\frac{2}{3} \\ U_{n+1}=\frac{1}{2} U_{n}+\frac{n}{2 \sqrt{2}}+\frac{1}{\sqrt{2}}\end{array}\right.
1. Calculer U1,U2U_{1}, U_{2} et U3U_{3}
2. On pose: n0Vn=Un2n\forall n \geq 0 \quad V_{n}=U_{n} \sqrt{2}-n. a. Calculer V0,V1V_{0}, V_{1} et V2V_{2} b. Montrer que (Vn)\left(V_{n}\right) est une suite géométrique c. Exprimer Vn\boldsymbol{V}_{\boldsymbol{n}} puis Un\boldsymbol{U}_{\boldsymbol{n}} en fonction de n\boldsymbol{n} d. Calculer en fonction de n:Sn=k=0k=nvkn: S_{n}=\sum_{k=0}^{k=n} v_{k}

See Solution

Problem 774

Problem A Continue the two sequences of numbers below and find an equation to calculate the nn-th value: \begin{tabular}{c|c|c|c|c|c|c|c|c} n\mathbf{n} & 1\mathbf{1} & 2\mathbf{2} & 3\mathbf{3} & 4\mathbf{4} & 5\mathbf{5} & 6\mathbf{6} & 7\mathbf{7} & Equation \\ \hlineana_{n} & 2 & 5 & 10 & 17 & 26 & 37 & & \\ \hlinebnb_{n} & 1 & 2 & 8 & 48 & 384 & 3840 & & \end{tabular}

See Solution

Problem 775

A new car is worth $25,000\$ 25,000. However, it loses 12%12 \% of its value each year due to depreciation. Write an explicit formula describing the value of the car, ana_{n}, after nn years.

See Solution

Problem 776

2. [-/1 Points] DETAILS MY NOTES SCALCET8M 11.8.042. 0/30 / 3 Submissions Used
Suppose that the radius of convergence of the power series cnxn\sum c_{n} x^{n} is RR. What is the radius of convergence of the power series cnx4n\sum c_{n} x^{4 n} ? \square

See Solution

Problem 777

12. Here are two sequencen: 1 EXAM
Sequence B a. For sequence AA, describe a way to produce each new term from the previous term. take Lle previas term and limes it by 10 b. For sequence B, describe a way to produce each new term from the previous term. c. Write a definition for the nnth term of sequence AA d. Write a definition for the nth term of sequence B e. If these sequences continue, then which is greater, A(6)A(6) or B(6)B(6) ? Explain or show how you know.

See Solution

Problem 778

A ball dropped from 256 feet bounces up 14\frac{1}{4} of its fall height. What height on the third bounce?

See Solution

Problem 779

 rem Complete the pattern: 4×4×=44×4×=4004×4×=4,0004×=40,0004×=400,0004\begin{array}{l}\text { rem Complete the pattern: } \\ \begin{array}{l} 4 \times \square \\ 4 \times=4 \\ 4 \times \\ 4 \times=400 \\ 4 \times \\ 4 \times=4,000 \\ 4 \times=40,000 \\ 4 \times=400,000 \\ 4\end{array}\end{array}

See Solution

Problem 780

um erro inferior a 10210^{-2}
4. Usando o método de ponto fixo, determinar o valor aproximado de 75\sqrt[5]{-7} com erro inferior a 10210^{-2}. 3.5

See Solution

Problem 781

1.2 Calculate m250(52 m)\sum_{\mathrm{m}-2}^{50}(5-2 \mathrm{~m})

See Solution

Problem 782

Дослідити на рівномірну збіжність: n=111+n2x2arctgxn,xR.\sum_{n=1}^{\infty} \frac{1}{1+n^{2} x^{2}} \operatorname{arctg} \frac{x}{n}, x \in \mathbb{R} .

See Solution

Problem 783

2) Theresa adds $3,000\$ 3,000 to her savings account on the first day of each year. Marcus adds $3,000\$ 3,000 to his savings account on the last day of each year. They both earn 7.5 percent annual interest. What is the difference in their savings account balances at the end of 34 years? You estimate that you will owe \$48,200 in student loans by the time you graduate. The interest rate is 6.52 percent. If you want to have this debt paid in full within six years, how much must you pay each month?

See Solution

Problem 784

Exercice ( 5 pts ) On considère la suite UU définie par {U0=23Un+1=3Un+22Un+3;nIN\left\{\begin{array}{c}U_{0}=\frac{2}{3} \\ U_{n+1}=\frac{3 U_{n}+2}{2 U_{n}+3} ; \forall n \in I N\end{array}\right.
1. Calculer U1;U2\boldsymbol{U}_{\mathbf{1}} ; \boldsymbol{U}_{\mathbf{2}}
2. Monter que nIN\forall n \in I N on a : 0Un10 \leq U_{n} \leq 1
3. On pose Vn=Un1NUn+1V_{n}=\frac{U_{n}-1^{N}}{U_{n}+1} a. Monter que la suite (Vn)\left(V_{n}\right) est une suite géométrique b. Calculer Vn\boldsymbol{V}_{\boldsymbol{n}} puis Un\boldsymbol{U}_{\boldsymbol{n}} en fonction de nn c. Calculer Sn=k=0nVnS_{n}=\sum_{k=0}^{n} V_{n} d. Calculer limVn;limUn\lim V_{n} ; \lim U_{n} et limSn\lim S_{n}

See Solution

Problem 785

Soit {Un}\left\{U_{n}\right\} et (Vn)\left(V_{n}\right) deux suites définies par: Un=2n+4n+32U_{n}=\frac{2^{n}+4 n+3}{2} et Vn=2n4n+32V_{n}=\frac{2^{n}-4 n+3}{2} On pose T1=Un+VnT_{1}=U_{n}+V_{n} et T2=UnVnT_{2}=U_{n}-V_{n} 1) Montrer que T1T_{1} est géométrique et que T2T_{2} est arithmétique ? 2) En déduire S1S_{1} et S2S_{2} en fonction de nn tels que: S1=K=0nUKS_{1}=\sum_{K=0}^{n} \boldsymbol{U}_{K} et S2=K=0nVKS_{2}=\sum_{K=0}^{n} V_{K}

See Solution
banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord