Question

Show that the coefficient of ${{x}^{n}}$ in the expansion of $\dfrac{x}{{{\left( 1-x \right)}^{2}}-cx}$ is $n\left\{ 1+\dfrac{{{n}^{2}}-1}{\left| \!{\underline {\,
  3 \,}} \right. }c+\dfrac{\left( {{n}^{2}}-1 \right)\left( {{n}^{2}}-4 \right)}{\left| \!{\underline {\,
  5 \,}} \right. }{{c}^{2}}+\dfrac{\left( {{n}^{2}}-1 \right)\left( {{n}^{2}}-4 \right)\left( {{n}^{2}}-9 \right)}{\left| \!{\underline {\,
  7 \,}} \right. }{{c}^{3}}+........ \right\}$

Answer 1

Hint: We can solve this question by using some of basic formula of infinite GP that is
$\dfrac{1}{1-x}=1+x+{{x}^{2}}+{{x}^{3}}+......$ where $-1\le x\le 1$ and another series ${{\left( 1-x \right)}^{-n}}=1+nx+\dfrac{n\left( n+1 \right)}{\left| \!{\underline {\,
  2 \,}} \right. }{{x}^{2}}+\dfrac{n\left( n+1 \right)\left( n+2 \right)}{\left| \!{\underline {\,
  3 \,}} \right. }......$ where $-1\le x\le 1$ we can write
${{\left( 1-x \right)}^{-n}}=1+\sum\limits_{r=0}^{\infty }{(n+r){{C}_{r+1}}}{{x}^{r+1}}$ . Coefficient of ${{x}^{r}}$ in ${{\left( 1-x \right)}^{-n}}$ is $\left( n+r-1 \right){{C}_{r}}$ .

Complete step by step answer:
 The coefficient of ${{x}^{n}}$ in the series $\dfrac{x}{{{\left( 1-x \right)}^{2}}-cx}$ is equal to coefficient of ${{x}^{n-1}}$ in the series $\dfrac{1}{{{\left( 1-x \right)}^{2}}-cx}$
Taking $\left( 1-{{x}^{2}} \right)$ common in the denominator we get
$\Rightarrow \dfrac{1}{{{\left( 1-x \right)}^{2}}-cx}=\dfrac{1}{{{\left( 1-x \right)}^{2}}}\left[ \dfrac{1}{1-\dfrac{cx}{{{\left( 1-x \right)}^{2}}}} \right]$
Let’ expand the term $\dfrac{1}{1-\left( \dfrac{cx}{{{\left( 1-x \right)}^{2}}} \right)}$
We can assume the $\dfrac{cx}{{{\left( 1-x \right)}^{2}}}$ as y then it will be $\dfrac{1}{1-y}$ then we can expand the term.

$\dfrac{1}{1-\left( \dfrac{cx}{{{\left( 1-x \right)}^{2}}} \right)}=1+\dfrac{cx}{{{\left( 1-x \right)}^{2}}}+{{\left( \dfrac{cx}{{{\left( 1-x \right)}^{2}}} \right)}^{2}}+{{\left( \dfrac{cx}{{{\left( 1-x \right)}^{2}}} \right)}^{3}}+......$ $-1\le \dfrac{cx}{{{\left( 1-x \right)}^{2}}}\le 1$

 $\Rightarrow \dfrac{1}{1-\left( \dfrac{cx}{{{\left( 1-x \right)}^{2}}} \right)}=1+\dfrac{cx}{{{\left( 1-x \right)}^{2}}}+\dfrac{{{c}^{2}}{{x}^{2}}}{{{\left( 1-x \right)}^{4}}}+\dfrac{{{c}^{3}}{{x}^{3}}}{{{\left( 1-x \right)}^{6}}}+.....$ $-1\le \dfrac{cx}{{{\left( 1-x \right)}^{2}}}\le 1$
Now we can write
$\dfrac{1}{{{\left( 1-x \right)}^{2}}-cx}=\dfrac{1}{{{\left( 1-x \right)}^{2}}}\left( 1+\dfrac{cx}{{{\left( 1-x \right)}^{2}}}+\dfrac{{{c}^{2}}{{x}^{2}}}{{{\left( 1-x \right)}^{4}}}+\dfrac{{{c}^{3}}{{x}^{3}}}{{{\left( 1-x \right)}^{6}}}+..... \right)$
$\dfrac{1}{{{\left( 1-x \right)}^{2}}-cx}=\left( \dfrac{1}{{{\left( 1-x \right)}^{2}}}+\dfrac{cx}{{{\left( 1-x \right)}^{4}}}+\dfrac{{{c}^{2}}{{x}^{2}}}{{{\left( 1-x \right)}^{6}}}+\dfrac{{{c}^{3}}{{x}^{3}}}{{{\left( 1-x \right)}^{8}}}+..... \right)$
So the coefficient of ${{x}^{n-1}}$ in $\dfrac{1}{{{\left( 1-x \right)}^{2}}-cx}$ is equal to coefficient of ${{x}^{n-1}}$ in $\left( \dfrac{1}{{{\left( 1-x \right)}^{2}}}+\dfrac{cx}{{{\left( 1-x \right)}^{4}}}+\dfrac{{{c}^{2}}{{x}^{2}}}{{{\left( 1-x \right)}^{6}}}+\dfrac{{{c}^{3}}{{x}^{3}}}{{{\left( 1-x \right)}^{8}}}+..... \right)$ = coefficient of ${{x}^{n-1}}$ in $\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}{{x}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}}$
=coefficient of ${{x}^{n-1-r}}$ in $\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}}$
We know that coefficient of ${{x}^{p}}$ in ${{\left( 1-x \right)}^{-m}}$ is $\left( m+p-1 \right){{C}_{p}}$ or $\left( m+p-1 \right){{C}_{m-1}}$
Here p is $n-1-r$ and m is $2r+2$
Coefficient of ${{x}^{n-1-r}}$ in $\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}}$=$\sum\limits_{r=0}^{\infty }{{}}$ (coefficient of ${{x}^{n-1-r}}$ in $\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}$ )

$\Rightarrow $ Coefficient of ${{x}^{n-1-r}}$ in $\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}}$ = $\sum\limits_{r=0}^{\infty }{{}}$${{c}^{r}}\times \left( n-1-r+2r+2-1 \right){{C}_{2r+2-1}}$
 $\Rightarrow $ Coefficient of ${{x}^{n-1-r}}$ in $\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}}$ = $\sum\limits_{r=0}^{\infty }{\left( n+r \right){{C}_{2r+1}}\times {{c}^{r}}}$
$\Rightarrow \sum\limits_{r=0}^{\infty }{\left( n+r \right){{C}_{2r+1}}\times {{c}^{r}}}=\sum\limits_{r=0}^{\infty }{\dfrac{\left( n+r \right)\times \left( n+r-1 \right)\times .........\left( n-(r-1) \right)\times \left( n-r \right)\times {{c}^{r}}}{2}}$
We can see there is $\left( n+r \right)\times \left( n-r \right)$ , $\left( n+r-1 \right)\times \left( n-(r-1) \right)$ ,……..till $\left( n+1 \right)\times \left( n-1 \right)$ and n is left out. We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ so taking n common and multiply these terms we get
$\Rightarrow \sum\limits_{r=0}^{\infty }{\dfrac{\left( n+r \right)\times \left( n+r-1 \right)\times .........\left( n-(r-1) \right)\times \left( n-r \right){{c}^{r}}}{\left| \!{\underline {\,
  2r+1 \,}} \right. }}=n\times \sum\limits_{r=0}^{\infty }{\dfrac{\left( {{n}^{2}}-{{r}^{2}} \right)\left( {{n}^{2}}-{{(r-1)}^{2}} \right).........\left( {{n}^{2}}-{{2}^{2}} \right)\left( {{n}^{2}}-{{1}^{2}} \right){{c}^{r}}}{\left| \!{\underline {\,
  2r+1 \,}} \right. }}$
 So expanding the term we get $n\left\{ 1+\dfrac{{{n}^{2}}-1}{\left| \!{\underline {\,
  3 \,}} \right. }c+\dfrac{\left( {{n}^{2}}-1 \right)\left( {{n}^{2}}-4 \right)}{\left| \!{\underline {\,
  5 \,}} \right. }{{c}^{2}}+\dfrac{\left( {{n}^{2}}-1 \right)\left( {{n}^{2}}-4 \right)\left( {{n}^{2}}-9 \right)}{\left| \!{\underline {\,
  7 \,}} \right. }{{c}^{3}}+........ \right\}$.
Now we can see from the above proof that the coefficient of
  ${{x}^{n}}$ in the expansion of $\dfrac{x}{{{\left( 1-x \right)}^{2}}-cx}$ is $n\left\{ 1+\dfrac{{{n}^{2}}-1}{\left| \!{\underline {\,
  3 \,}} \right. }c+\dfrac{\left( {{n}^{2}}-1 \right)\left( {{n}^{2}}-4 \right)}{\left| \!{\underline {\,
  5 \,}} \right. }{{c}^{2}}+\dfrac{\left( {{n}^{2}}-1 \right)\left( {{n}^{2}}-4 \right)\left( {{n}^{2}}-9 \right)}{\left| \!{\underline {\,
  7 \,}} \right. }{{c}^{3}}+........ \right\}$

Keep in mind that all the expansion of series is valid when
$-1\le x\le 1$
$-1\le \dfrac{cx}{{{\left( 1-x \right)}^{2}}}\le 1$

Note: Always keep in mind that coefficient of ${{x}^{n}}$ in in a series which is in the form ${{x}^{r}}\times S$ where S is a series will be equal to coefficient of ${{x}^{n-r}}$ in S. we have seen this in the above question. Some infinite series will be valid only in a specific range of x. So try to mention the range while solving.