Solution: 2014 Winter Final - 3

Author: Michiel Smid

Question

What is $$ \sum_{k=0}^{45} {45 \choose k} (-3)^{2k}. $$

(a)

$(-8)^{45}$

(b)

$4^{45}$

(c)

$(-2)^{45}$

(d)

$10^{45}$

Let’s break it down

$ \sum_{k=0}^{45} \binom{45}{k} {{(-3)}^{2k}} $

$ = \sum_{k=0}^{45} \binom{45}{k} {9}^{k} $

$ = \sum_{k=0}^{45} \binom{45}{k} {9}^{k} {1}^{45-k}$

$ = {(1+9)}^{45} $

$ = {10}^{45} $