If $m,n$ are positive integers and $$(1+ x + x^2 + x^3... + x^n)^m = a + a'x + a''x^2 + a'''x^3...$$ Prove that the value of $$a' + 3a'' + 6a''' + 10a'''' ...= \frac{mn(2m+1)^n(3mn+n-2)}{6}$$ I tried to open the the given polynomial as a GP and expressing coefficient of any general term, and then summing the term required for expression but was unable to reach answer.
Any alternate way or any improvisation?
(Let us call $a^{i}{'}=a_i$)
Hint : $$\text{Your sum is}~:~\sum \Big(\frac{n^2-n}{2}\Big)a_n=\sum \frac{n^2}{2}a_n-\sum \frac{n}{2}a_n$$
For $\sum n^2a_n$ : Differentiate both the sides, then multiply by $x$, differentiate again , then put $x=1$
For $\sum na_n$ : Differentiate and put $x=1$
Can you procced now?