What is the sum of $1^3q + 2^3q^3 + 3^3q^3 +\cdots+ n^3q^n$?

Question

What is the sum of $1^3*q + 2^3*q^2 + 3^3*q^3 +...+ n^3*q^n ?$

Answer 1

Hint. Summing a GP, $$q+q^2+q^3+\cdots+q^n=q\frac{q^n-1}{q-1}=\frac{q^{n+1}-q}{q-1}\ .$$ Differentiating both sides with respect to $q$, $$1+2q+3q^2+\cdots+nq^{n-1}=\{\hbox{something}\}\ .$$ Multiplying by $q$ gives $$q+2q^2+3q^3+\cdots+nq^n=\{\hbox{something else}\}\ .$$ Differentiating again, $$1+2^2q+3^2q^2+\cdots+n^2q^{n-1}=\cdots\ .$$ Keep going in the same way.

Answer 2

Hint:

You are, unless I am mistaken, asking for $$ 1^3q+2^3q^2+\cdots+n^3q^n=\sum_{i=1}^{n}i^3q^i. $$ Note that $$ \sum_{i=1}^{n}i^3q^i=q\sum_{i=1}^{n}i^3q^{i-1}=q\sum_{i=1}^{n}i^2\cdot iq^{i-1}=q\cdot\frac{d}{dq}\left[\sum_{i=1}^{n}i^2q^i\right]. $$ Repeating this trick, you can write $\sum i^2q^i$ in terms of $\sum iq^i$, and $\sum iq^i$ in terms of $\sum q^i$. And THAT one, we have a nice and easy formula for.