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

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?

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 $$V_n = \frac{1}{10} n (n + 1) (n + 2) (n + 3) (2 n + 3)$$
• I got $V_n = S_n$ by using power series expansion. As $T_n$ is of order $n^4$, $S_n$ has to be of order $n^5$. You may therefore also set a general formula $V_n = a_0 + a_1 n + a_2 n^2 + a_3 n^3 + a_4 n^4 + a_5 n^5$ and determine the coefficients $a_i$ by setting all powers $n^k$ equal in $V_n - V_{n-1} = T_n$. – Andreas Jan 27 at 19:46
• I've laid out another path to arrive at $V_n$ - with unknown coefficients to be determined - in my previous comment. Kindly use that one. – Andreas Jan 28 at 8:15