Find the coefficient of $x^n$ in $(x^2 +x^3 +x^4 +\cdots)^5$

Question

I have got stuck on this question, though I realise that I have probably got really close to an answer. This is how I approached it: \begin{align*}f(x) &= (x^2+x^3+x^4+\cdots)^5\\ &= x^{10}(1 + x + x^2 + \cdots)^5\\ &= x^{10}\left(\frac{1-x^{m}}{1-x}\right)^5\\ &= x^{10}(1-x^{m})^5(1-x)^{-5}.\end{align*} Then I have used the binomial theorem: \begin{align*}(1-x^{m})^5 &= \sum^5_{i=0}\binom{5}{i}(-1)^i(x^m)^i,\\ (1-x)^{-5} &= \sum^\infty_{j=0}\binom{4+j}{j}(x)^j.\end{align*} Therefore, $$f(x) = x^{10}\cdot\left(\sum^5_{i=0}\binom{5}{i}(-1)^i(x^m)^i\right)\cdot\left(\sum^\infty_{j=0}\binom{4+j}{j}x^j\right),$$ and as I work out coefficient of $x^n$ I arrive to: $$[x^n] = \binom{5}{0} \binom{n-6}{n-10} $$ I think my answer is incomplete and unfortunately, I don't have a solution to check it. I would really appreciate your help. Thank you

Answer 1

As you had before, we have $$f(x)=(x^2+x^3+\ldots)^5$$ $$f(x)=x^{10}(1+x+\ldots)^5$$ $$f(x)=\frac{x^{10}}{(1-x)^5}$$ $$f(x)=\frac{x^{10}}{(1-x)^5}$$ $$f(x)=x^{10}\sum_{k=0}^\infty \binom{k+4}{4}x^k$$ $$f(x)=\sum_{k=0}^\infty \binom{k+4}{4}x^{k+10}$$ Hence, the coefficient of $x^n$ is $\binom{n-6}{4}$