Finding the coefficient of $x^m$ in $(x+x^2+x^3+.....+x^i)^n$ EDIT:
My doubt may be silly but if the expansion of $(1-x)^{-n}$ is infinite, how come wolfram displayed 11 terms?
Wolfram result
Can anybody help to solve the whole equation?

I'm trying to solve this question.
I have an equation $(x+x^2+x^3+x^4+x^5+x^6)^2$  [ Taken 6-sided die x 2].
I want to find the coefficient of $x^5$.
I have reduced the equation to $x^2(1-x^6)^2(1-x)^{-2}$.
For $(1-x)^{-2}$,
$$(1-x)^{-2}=\sum_{k\ge0}(-1)^k\binom{-2}{k}x^k=\sum_{k\ge0}\binom{2+k-1}{k}x^k $$
I got  -> $1+2x+3x^2$
For $(1-x^6)^2$
$$(1-x^6)^2=\sum_{k\ge0}(-1)^k\binom{2}{k}x^{6k}$$  
I got -> $1-2x^6+x^{12}$
So my equation will be$$x^2(1+2x+3x^2)(1-2x^6+x^{12})$$
Am I correct  till here? Because when I multiply all the terms I'm getting
$$x^2+2x^3+3x^4-2x^8-4x^9-6x^{10}+x^{14}+2x^{15}+3x^{16}$$ 
Where am I going wrong?

This GitHub repository solved my problem.
