Jaqueline Kirby

Answered

2022-06-25

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

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:

$\begin{array}{rl}f(x)& =({x}^{2}+{x}^{3}+{x}^{4}+\cdots {)}^{5}\\ & ={x}^{10}(1+x+{x}^{2}+\cdots {)}^{5}\\ & ={x}^{10}{\left(\frac{1-{x}^{m}}{1-x}\right)}^{5}\\ & ={x}^{10}(1-{x}^{m}{)}^{5}(1-x{)}^{-5}.\end{array}$

Then I have used the binomial theorem:

$\begin{array}{rl}(1-{x}^{m}{)}^{5}& =\sum _{i=0}^{5}{\textstyle (}\genfrac{}{}{0ex}{}{5}{i}{\textstyle )}(-1{)}^{i}({x}^{m}{)}^{i},\\ (1-x{)}^{-5}& =\sum _{j=0}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{4+j}{j}{\textstyle )}(x{)}^{j}.\end{array}$

Therefore, $f(x)={x}^{10}\cdot (\sum _{i=0}^{5}{\textstyle (}\genfrac{}{}{0ex}{}{5}{i}{\textstyle )}(-1{)}^{i}({x}^{m}{)}^{i})\cdot \left(\sum _{j=0}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{4+j}{j}{\textstyle )}{x}^{j}\right),$

and as I work out coefficient of ${x}^{n}$ I arrive to:

$[{x}^{n}]={\textstyle (}\genfrac{}{}{0ex}{}{5}{0}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{n-6}{n-10}{\textstyle )}$

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

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:

$\begin{array}{rl}f(x)& =({x}^{2}+{x}^{3}+{x}^{4}+\cdots {)}^{5}\\ & ={x}^{10}(1+x+{x}^{2}+\cdots {)}^{5}\\ & ={x}^{10}{\left(\frac{1-{x}^{m}}{1-x}\right)}^{5}\\ & ={x}^{10}(1-{x}^{m}{)}^{5}(1-x{)}^{-5}.\end{array}$

Then I have used the binomial theorem:

$\begin{array}{rl}(1-{x}^{m}{)}^{5}& =\sum _{i=0}^{5}{\textstyle (}\genfrac{}{}{0ex}{}{5}{i}{\textstyle )}(-1{)}^{i}({x}^{m}{)}^{i},\\ (1-x{)}^{-5}& =\sum _{j=0}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{4+j}{j}{\textstyle )}(x{)}^{j}.\end{array}$

Therefore, $f(x)={x}^{10}\cdot (\sum _{i=0}^{5}{\textstyle (}\genfrac{}{}{0ex}{}{5}{i}{\textstyle )}(-1{)}^{i}({x}^{m}{)}^{i})\cdot \left(\sum _{j=0}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{4+j}{j}{\textstyle )}{x}^{j}\right),$

and as I work out coefficient of ${x}^{n}$ I arrive to:

$[{x}^{n}]={\textstyle (}\genfrac{}{}{0ex}{}{5}{0}{\textstyle )}{\textstyle (}\genfrac{}{}{0ex}{}{n-6}{n-10}{\textstyle )}$

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 & Explanation

jmibanezla

Expert

2022-06-26Added 17 answers

Step 1

As you had before, we have

$f(x)=({x}^{2}+{x}^{3}+\dots {)}^{5}$

$f(x)={x}^{10}(1+x+\dots {)}^{5}$

$f(x)=\frac{{x}^{10}}{(1-x{)}^{5}}$

$f(x)=\frac{{x}^{10}}{(1-x{)}^{5}}$

$f(x)={x}^{10}\sum _{k=0}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{k+4}{4}{\textstyle )}{x}^{k}$

$f(x)=\sum _{k=0}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{k+4}{4}{\textstyle )}{x}^{k+10}$

Step 2

Hence, the coefficient of ${x}^{n}$ is $(}\genfrac{}{}{0ex}{}{n-6}{4}{\textstyle )$.

As you had before, we have

$f(x)=({x}^{2}+{x}^{3}+\dots {)}^{5}$

$f(x)={x}^{10}(1+x+\dots {)}^{5}$

$f(x)=\frac{{x}^{10}}{(1-x{)}^{5}}$

$f(x)=\frac{{x}^{10}}{(1-x{)}^{5}}$

$f(x)={x}^{10}\sum _{k=0}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{k+4}{4}{\textstyle )}{x}^{k}$

$f(x)=\sum _{k=0}^{\mathrm{\infty}}{\textstyle (}\genfrac{}{}{0ex}{}{k+4}{4}{\textstyle )}{x}^{k+10}$

Step 2

Hence, the coefficient of ${x}^{n}$ is $(}\genfrac{}{}{0ex}{}{n-6}{4}{\textstyle )$.

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