Math  /  Algebra

Question3. 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: \qquad

Studdy Solution
Podstawiamy S(N)=N(N+1)2=55 S(N) = \frac{N(N+1)}{2} = 55 i rozwiązujemy dla N N :
N(N+1)2=55 \frac{N(N+1)}{2} = 55
Pomnóżmy obie strony przez 2:
N(N+1)=110 N(N+1) = 110
Rozwiązujemy równanie kwadratowe:
N2+N110=0 N^2 + N - 110 = 0
Używamy wzoru kwadratowego:
N=b±b24ac2a N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
gdzie 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}
Wybieramy dodatnie rozwiązanie:
N=202=10 N = \frac{20}{2} = 10
Wartość N N to:
10 \boxed{10}

View Full Solution - Free
Was this helpful?

Studdy solves anything!

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