Suppose the sum \[ \sum_{k = 0}^{49} ( - 1)^k \binom{99}{2k}, \] where \binom{n}{j} = \frac {n! }{j! (n - j)! }, is written in the form a^b, where a and b are integers and b is as large as possible. find a+b.

your answer would be b

step-by-step explanation:

step-by-step explanation:

i have done this all on a piece of paper but idk how to put it on here cuz it needs to be drawn

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58