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Math  /  Algebra

Question3. IE S(N)=∑k=1NS(N)=\sum_{k=1}^{N} \& then which value of NN solves the following equation? ∑n=1SN14n=43(455−1).\sum_{n=1}^{S N 1} 4^{n}=\frac{4}{3}\left(4^{55}-1\right) .
ANS: value of

Studdy Solution

STEP 1

1. S(N) S(N) is defined as the sum of the first N N natural numbers, i.e., S(N)=∑k=1Nk=N(N+1)2 S(N) = \sum_{k=1}^{N} k = \frac{N(N+1)}{2} .
2. We need to find N N such that the sum of powers of 4 from 1 to S(N) S(N) equals the given expression.

STEP 2

1. Express S(N) S(N) in terms of N N .
2. Use the formula for the sum of a geometric series to express the left side of the equation.
3. Equate the expressions and solve for N N .

STEP 3

Express S(N) S(N) in terms of N N :
S(N)=N(N+1)2 S(N) = \frac{N(N+1)}{2}

STEP 4

Use the formula for the sum of a geometric series. The sum of the series ∑n=1m4n \sum_{n=1}^{m} 4^n is given by:
∑n=1m4n=44m−14−1=43(4m−1) \sum_{n=1}^{m} 4^n = 4 \frac{4^m - 1}{4 - 1} = \frac{4}{3}(4^m - 1)
Set m=S(N) m = S(N) , so:
∑n=1S(N)4n=43(4S(N)−1) \sum_{n=1}^{S(N)} 4^n = \frac{4}{3}(4^{S(N)} - 1)

STEP 5

Equate the expressions from the problem and solve for S(N) S(N) :
43(4S(N)−1)=43(455−1) \frac{4}{3}(4^{S(N)} - 1) = \frac{4}{3}(4^{55} - 1)
Since the coefficients are the same, equate the exponents:
4S(N)=455 4^{S(N)} = 4^{55}
This implies:
S(N)=55 S(N) = 55

STEP 6

Now solve for N N using the expression for S(N) S(N) :
N(N+1)2=55 \frac{N(N+1)}{2} = 55
Multiply both sides by 2:
N(N+1)=110 N(N+1) = 110
Rearrange into a quadratic equation:
N2+N−110=0 N^2 + N - 110 = 0
Solve this quadratic equation using the quadratic formula:
N=−b±b2−4ac2a N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=1 a = 1 , b=1 b = 1 , c=−110 c = -110 :
N=−1±1+4402 N = \frac{-1 \pm \sqrt{1 + 440}}{2} N=−1±4412 N = \frac{-1 \pm \sqrt{441}}{2} N=−1±212 N = \frac{-1 \pm 21}{2}
This gives two possible solutions:
N=202=10 N = \frac{20}{2} = 10 N=−222=−11 N = \frac{-22}{2} = -11
Since N N must be a positive integer, we have:
N=10 N = 10
The value of N N is:
10 \boxed{10}

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