Question

Someone check my summation please?

Answer 1

\[\large \sum_{n=2}^{\infty}n(n-1)a_nx^n=\sum_{n=\color{red}{0}}^{\infty}n(n-1)a_nx^n\]
and
\[\large \sum_{n=1}^{\infty}na_nx^n=\sum_{n=\color{red}{0}}^{\infty}na_nx^n\]
Right? It's part of a differential equation problem, but I'm just checking to see if I can make this simplification

Answer 2

right at n=0, n(n-1)a_nx^n is 0
and
at n=1, n(n-1)a_nx^n is also 0
so yep you can start at n=2 or n=0 for the first one

also similar thing with the second equality

Answer 3

Sweet, thanks! X)