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(4551)\sum_{n=1}^{S(N)} 4^{n}=\frac{4}{3}\left(4^{55}-1\right)

ANS: N=10N=10

Studdy Solution
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+N110=0 N^2 + N - 110 = 0
Using the quadratic formula N=b±b24ac2a 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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