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Question3. If S(N)=∑k=1NkS(N)=\sum_{k=1}^{N} k then which value ∑n=1S(N)4n=43(455−1)\sum_{n=1}^{S(N)} 4^{n}=\frac{4}{3}\left(4^{55}-1\right)

Studdy Solution

STEP 1

1. S(N)=∑k=1Nk S(N) = \sum_{k=1}^{N} k is the sum of the first N N natural numbers.
2. The formula for the sum of the first N N natural numbers is S(N)=N(N+1)2 S(N) = \frac{N(N+1)}{2} .
3. The given equation involves a geometric series with a common ratio of 4.

STEP 2

1. Calculate S(N) S(N) using the formula for the sum of the first N N natural numbers.
2. Set up the equation for the sum of the geometric series.
3. Solve for N N using the given equation.

STEP 3

Calculate S(N) S(N) using the formula for the sum of the first N N natural numbers:
S(N)=N(N+1)2 S(N) = \frac{N(N+1)}{2}

STEP 4

The problem states:
∑n=1S(N)4n=43(455−1) \sum_{n=1}^{S(N)} 4^n = \frac{4}{3}(4^{55} - 1)
This is a geometric series with the first term a=4 a = 4 and the common ratio r=4 r = 4 .
The formula for the sum of the first m m terms of a geometric series is:
Sm=arm−1r−1 S_m = a \frac{r^m - 1}{r - 1}
In this case, a=4 a = 4 , r=4 r = 4 , and m=S(N) m = S(N) .

STEP 5

Set up the equation for the sum of the geometric series:
44S(N)−14−1=43(455−1) 4 \frac{4^{S(N)} - 1}{4 - 1} = \frac{4}{3}(4^{55} - 1)
Simplify the left side:
43(4S(N)−1)=43(455−1) \frac{4}{3}(4^{S(N)} - 1) = \frac{4}{3}(4^{55} - 1)

STEP 6

Since the coefficients 43\frac{4}{3} are the same on both sides, we equate the exponents:
4S(N)−1=455−1 4^{S(N)} - 1 = 4^{55} - 1
This implies:
S(N)=55 S(N) = 55

STEP 7

Now solve for N N using the equation S(N)=N(N+1)2=55 S(N) = \frac{N(N+1)}{2} = 55 :
N(N+1)2=55 \frac{N(N+1)}{2} = 55
Multiply both sides by 2 to eliminate the fraction:
N(N+1)=110 N(N+1) = 110

STEP 8

Solve the quadratic equation:
N2+N−110=0 N^2 + N - 110 = 0
Use 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 , and 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}
The solutions are:
N=202=10andN=−222=−11 N = \frac{20}{2} = 10 \quad \text{and} \quad 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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