Solution: 2014 Fall Final - 3

Author: Michiel Smid

Question

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

(a)

$(-2)^{45}$

(b)

$(-8)^{45}$

(c)

$10^{45}$

(d)

$4^{45}$

$ \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} $