Sum of series: $1*3*(2^2) + 2*4*(3^2) + 3*5*(4^2) + \dots$?

Question

I am trying to find the sum of the above series.

The sum till n terms can be found using power series expansion. However, I'm trying to solve this using the method of difference (a.k.a. Telescoping sum or $V_n$ method).

In this method, the general term is expressed as the difference between two consecutive values of some function. Like the following:

$T_n = V_n - V_{n-1}$

and then the sum is taken which comes to be

$S_n = V_n - V_0$

The general term of the series in question can be represented as a product:

$T_n = n(n+1)^2(n+2)$

But I am unable to represent this as a difference. How can I proceed from here to find the sum?

Answer 1

Since $V_n = S_n + V_0$ you have that $V_n - V_{n-1} = S_n - S_{n-1}$ so your search for $V_n$ is identical to finding $S_n$, besides a free constant.

So you may well set $V_n$ to the (known or guessed) sum $S_n$ and write $$V_n = \frac{1}{10} n (n + 1) (n + 2) (n + 3) (2 n + 3)$$