<s>[INST] Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$ Find the sum of $n$ terms of following series:

$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$

I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way find the sum?
 [/INST] The answer is $S_n = \frac{1}{5}n^5 + \frac{3}{2}n^4 + 4n^3 + \frac{9}{2}n^2 + \frac{54}{30}n $
$\sum n^4 $ is not really a problem.
You can do it using this method here 
https://www.coursera.org/learn/calculus1/lecture/uR7YK/what-is-the-sum-of-n-4-for-n-1-to-n-k
I don't know if this is helpful but this is how i got it
$S_n = \sum a_n$
$a_n = n(2+n)(1+n)^2 = n^4+4n^3+5n^2+2n$
therefore : $ S_n = \sum {(n^4+4n^3+5n^2+2n)} $
$S_n = \sum n^4 + 4\sum n^3 + 5\sum n^2 + 2\sum n $
$S_n = \frac{1}{30}( 6n^5 +15n^4 + 10n^3- n) +4(\frac{1}{4}(n^4+2n^3+n^2) + 5(\frac{1}{6}(2n^3 + 3n^2 +n ) ) + 2(\frac{1}{2}(n^2 +n ) ) $
hence, $S_n = \frac{1}{5}n^5 + \frac{3}{2}n^4 + 4n^3 + \frac{9}{2}n^2 + \frac{54}{30}n $
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