∑ n = 0 ∞ n 3 ( ( 2 n ) ! ) + ( 2 n - 1 ) ( n ! ) ( n ! ) ( ( 2 n ) ! ) = a e + b e + c where a , b , c ∈ ℤ and e = ∑ n = 0 ∞ 1 n ! Then a 2 - b + c is equal to _______

Step-by-step solution

Given: ∑ n = 0 ∞ n 3 2 n ! + 2 n - 1 n ! n ! × 2 n ! = ∑ n = 0 ∞ n 3 2 n ! n ! × 2 n ! + ∑ n = 0 ∞ 2 n - 1 n ! n ! × 2 n ! = ∑ n = 0 ∞ n 3 n ! + ∑ n = 0 ∞ 2 n - 1 2 n ! = ∑ n = 0 ∞ n n - 1 n - 2 + 3 n n - 1 + n n ! + ∑ n = 1 ∞ 1 2 n - 1 ! - ∑ n = 0 ∞ 1 2 n ! = ∑ n = 0 ∞ n n - 1 n - 2 n ! + ∑ n = 0 ∞ 3 n n - 1 n ! + ∑ n = 0 ∞ n n ! + e - e - 1 2 - e + e - 1 2 = ∑ n