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Math  /  Algebra

QuestionANS: x∈(1,5)∪18;+x \in(1,5) \cup 18 ;+
3. If S(N)=∑k=1NkS(N)=\sum_{k=1}^{N} k then which value of NN solves the following equation? ∑n=1S(N)4n=43(455−1)\sum_{n=1}^{S(N)} 4^{n}=\frac{4}{3}\left(4^{55}-1\right)
ANS: N=10N=10

Studdy Solution

STEP 1

1. The function S(N)=∑k=1Nk S(N) = \sum_{k=1}^{N} k represents the sum of the first N N natural numbers.
2. The equation involves a geometric series on the left-hand side.
3. We need to find the value of N N such that the sum of powers of 4 equals the given expression.

STEP 2

1. Express S(N) S(N) in terms of N N .
2. Recognize and simplify the geometric series.
3. Solve for N N .

STEP 3

First, express S(N) S(N) in terms of N N . The sum of the first N N natural numbers is given by:
S(N)=N(N+1)2 S(N) = \frac{N(N+1)}{2}

STEP 4

Recognize that the left-hand side of the equation is a geometric series:
∑n=1S(N)4n=4+42+43+…+4S(N) \sum_{n=1}^{S(N)} 4^n = 4 + 4^2 + 4^3 + \ldots + 4^{S(N)}
The sum of a geometric series is given by:
Sum=arn−1r−1 \text{Sum} = a \frac{r^n - 1}{r - 1}
where a=4 a = 4 , r=4 r = 4 , and n=S(N) n = S(N) . So we have:
∑n=1S(N)4n=44S(N)−14−1=4S(N)+1−43 \sum_{n=1}^{S(N)} 4^n = 4 \frac{4^{S(N)} - 1}{4 - 1} = \frac{4^{S(N)+1} - 4}{3}

STEP 5

Set the geometric series equal to the given expression and solve for S(N) S(N) :
4S(N)+1−43=43(455−1) \frac{4^{S(N)+1} - 4}{3} = \frac{4}{3}(4^{55} - 1)
Multiply both sides by 3 to eliminate the fraction:
4S(N)+1−4=4(455−1) 4^{S(N)+1} - 4 = 4(4^{55} - 1)
Simplify the right-hand side:
4S(N)+1−4=456−4 4^{S(N)+1} - 4 = 4^{56} - 4
Since the terms involving −4-4 cancel out, we equate the exponents:
S(N)+1=56 S(N)+1 = 56
S(N)=55 S(N) = 55

STEP 6

Substitute S(N)=55 S(N) = 55 into 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
Solve the quadratic equation:
N2+N−110=0 N^2 + N - 110 = 0
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 , 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 positive solution is:
N=202=10 N = \frac{20}{2} = 10
The value of N N is:
10 \boxed{10}

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