So many numbers!

Algebra Level 3

k = 0 4999 2 2 k = a b 1 c \large \sum_{k = 0}^{4999}{2^{2k}} = \dfrac{a^b - 1}{c} In the equation above, a a , b b , and c c are positive integers where a a and c c are prime numbers.

Find a + b + c a+b+c .


The answer is 10005.

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1 solution

Zach Abueg
Aug 17, 2017

Relevant wiki: Geometric Progression Sum

S = k = 0 4999 2 2 k = k = 0 4999 4 k S n = a ( r n + 1 1 ) r 1 = 4 5000 1 4 1 = 2 10000 1 3 \displaystyle \begin{aligned} S & = \sum_{k = 0}^{4999} 2^{2k} \\ & = \sum_{k = 0}^{4999} 4^k & \small \color{#3D99F6} S_n = \frac{a\left(r^{n + 1} - 1\right)}{r - 1} \\ & = \frac{4^{5000} - 1}{4 - 1} \\ & = \frac{2^{10000} - 1}{3} \end{aligned}

a + b + c = 2 + 10000 + 3 = 10005 \displaystyle \implies a + b + c = 2 + 10000 + 3 = \boxed{10005}

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