The sum of the infinite series _ n=1 ^ n^2 2^n n! is equal to

Step-by-step solution

Let S = _ n=1 ^ n^2 2^n n! Since n! = n (n-1)! , we can cancel one factor of n for n 1 : S = _ n=1 ^ n 2^n (n-1)! Rewrite n as (n-1) + 1 to split the fraction: S = _ n=1 ^ (n-1+1) 2^n (n-1)! S = _ n=2 ^ (n-1) 2^n (n-1)! + _ n=1 ^ 2^n (n-1)! Cancel (n-1) in the first term: S = _ n=2 ^ 2^n (n-2)! + _ n=1 ^ 2^n (n-1)! To use the standard exponential series e^x = _ r=0 ^ x^r r! , we adjust the powers of 2 to match the factorials in the denominators: S = 2^2 _ n=2 ^ 2^ n-2 (n-2)! + 2^1 _ n=1 ^ 2^ n-1 (n-1)! Let r = n-2