 Jaqueline Kirby

2022-06-25

Find the coefficient of ${x}^{n}$ in $\left({x}^{2}+{x}^{3}+{x}^{4}+\cdots {\right)}^{5}$.
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:
$\begin{array}{rl}f\left(x\right)& =\left({x}^{2}+{x}^{3}+{x}^{4}+\cdots {\right)}^{5}\\ & ={x}^{10}\left(1+x+{x}^{2}+\cdots {\right)}^{5}\\ & ={x}^{10}{\left(\frac{1-{x}^{m}}{1-x}\right)}^{5}\\ & ={x}^{10}\left(1-{x}^{m}{\right)}^{5}\left(1-x{\right)}^{-5}.\end{array}$
Then I have used the binomial theorem:
$\begin{array}{rl}\left(1-{x}^{m}{\right)}^{5}& =\sum _{i=0}^{5}\left(\genfrac{}{}{0}{}{5}{i}\right)\left(-1{\right)}^{i}\left({x}^{m}{\right)}^{i},\\ \left(1-x{\right)}^{-5}& =\sum _{j=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{4+j}{j}\right)\left(x{\right)}^{j}.\end{array}$
Therefore, $f\left(x\right)={x}^{10}\cdot \left(\sum _{i=0}^{5}\left(\genfrac{}{}{0}{}{5}{i}\right)\left(-1{\right)}^{i}\left({x}^{m}{\right)}^{i}\right)\cdot \left(\sum _{j=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{4+j}{j}\right){x}^{j}\right),$
and as I work out coefficient of ${x}^{n}$ I arrive to:
$\left[{x}^{n}\right]=\left(\genfrac{}{}{0}{}{5}{0}\right)\left(\genfrac{}{}{0}{}{n-6}{n-10}\right)$
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 jmibanezla

Expert

Step 1
As you had before, we have
$f\left(x\right)=\left({x}^{2}+{x}^{3}+\dots {\right)}^{5}$
$f\left(x\right)={x}^{10}\left(1+x+\dots {\right)}^{5}$
$f\left(x\right)=\frac{{x}^{10}}{\left(1-x{\right)}^{5}}$
$f\left(x\right)=\frac{{x}^{10}}{\left(1-x{\right)}^{5}}$
$f\left(x\right)={x}^{10}\sum _{k=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{k+4}{4}\right){x}^{k}$
$f\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0}{}{k+4}{4}\right){x}^{k+10}$
Step 2
Hence, the coefficient of ${x}^{n}$ is $\left(\genfrac{}{}{0}{}{n-6}{4}\right)$.

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