How do you perform this operation? #(x^2+2)sum_(n=2)^oo(n(n-1)a_nx^(n-2))#
1 Answer
Oct 30, 2017
# (x^2+2)sum_(n=2)^oo(n(n-1)a_nx^(n-2)) = sum_(n=2)^oo n(n-1)a_n(x^n + x^(n-2)) #
Explanation:
We seek to simplify:
# S = (x^2+2)sum_(n=2)^oo(n(n-1)a_nx^(n-2)) #
The outer terms are independent of the summation index, so we have:
# S = x^2sum_(n=2)^oo n(n-1)a_nx^(n-2) + 2sum_(n=2)^oo n(n-1)a_nx^(n-2) #
# \ \ = sum_(n=2)^oo n(n-1)a_nx^2x^(n-2) + 2sum_(n=2)^oo n(n-1)a_nx^(n-2) #
# \ \ = sum_(n=2)^oo n(n-1)a_nx^n + 2sum_(n=2)^oo n(n-1)a_nx^(n-2) #
# \ \ = sum_(n=2)^oo n(n-1)a_n(x^n + x^(n-2)) #
