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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?

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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 $$V_n = \frac{1}{10} n (n + 1) (n + 2) (n + 3) (2 n + 3)$$

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  • $\begingroup$ I didn't understand what you mean by setting Vn to the sum. I don't know the sum and can't guess it. Also, how did you arrive at the last equation? Can you please elaborate more? $\endgroup$ – user638500 Jan 27 at 15:25
  • $\begingroup$ 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 $V_n - V_{n-1} = T_n$. $\endgroup$ – Andreas Jan 27 at 19:46
  • $\begingroup$ I have mentioned in the question that I don't want to use Power series expansion. I am in high school and don't know calculus or Power series expansion. However, thanks for the answer. $\endgroup$ – user638500 Jan 28 at 3:12
  • $\begingroup$ 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. $\endgroup$ – Andreas Jan 28 at 8:15

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