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Question: Show that the coefficient of \({{x}^{n}}\) in the expansion of \(\dfrac{x}{{{\left( 1-x \right)}^{2}...

Show that the coefficient of xn{{x}^{n}} in the expansion of x(1−x)2−cx\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\\}

Explanation

Solution

We can solve this question by using some of basic formula of infinite GP that is
11−x=1+x+x2+x3+......\dfrac{1}{1-x}=1+x+{{x}^{2}}+{{x}^{3}}+...... where −1≤x≤1-1\le x\le 1 and another series (1−x)−n=1+nx+n(n+1)∣! 2 ‾x2+n(n+1)(n+2)∣! 3 ‾......{{\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≤x≤1-1\le x\le 1 we can write
(1−x)−n=1+∑r=0∞(n+r)Cr+1xr+1{{\left( 1-x \right)}^{-n}}=1+\sum\limits_{r=0}^{\infty }{(n+r){{C}_{r+1}}}{{x}^{r+1}} . Coefficient of xr{{x}^{r}} in (1−x)−n{{\left( 1-x \right)}^{-n}} is (n+r−1)Cr\left( n+r-1 \right){{C}_{r}} .

Complete step by step answer:
The coefficient of xn{{x}^{n}} in the series x(1−x)2−cx\dfrac{x}{{{\left( 1-x \right)}^{2}}-cx} is equal to coefficient of xn−1{{x}^{n-1}} in the series 1(1−x)2−cx\dfrac{1}{{{\left( 1-x \right)}^{2}}-cx}
Taking (1−x2)\left( 1-{{x}^{2}} \right) common in the denominator we get
⇒1(1−x)2−cx=1(1−x)2[11−cx(1−x)2]\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 11−(cx(1−x)2)\dfrac{1}{1-\left( \dfrac{cx}{{{\left( 1-x \right)}^{2}}} \right)}
We can assume the cx(1−x)2\dfrac{cx}{{{\left( 1-x \right)}^{2}}} as y then it will be 11−y\dfrac{1}{1-y} then we can expand the term.

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

⇒11−(cx(1−x)2)=1+cx(1−x)2+c2x2(1−x)4+c3x3(1−x)6+.....\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≤cx(1−x)2≤1-1\le \dfrac{cx}{{{\left( 1-x \right)}^{2}}}\le 1
Now we can write
1(1−x)2−cx=1(1−x)2(1+cx(1−x)2+c2x2(1−x)4+c3x3(1−x)6+.....)\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)
1(1−x)2−cx=(1(1−x)2+cx(1−x)4+c2x2(1−x)6+c3x3(1−x)8+.....)\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 xn−1{{x}^{n-1}} in 1(1−x)2−cx\dfrac{1}{{{\left( 1-x \right)}^{2}}-cx} is equal to coefficient of xn−1{{x}^{n-1}} in (1(1−x)2+cx(1−x)4+c2x2(1−x)6+c3x3(1−x)8+.....)\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 xn−1{{x}^{n-1}} in ∑r=0∞crxr(1−x)2r+2\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}{{x}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}}
=coefficient of xn−1−r{{x}^{n-1-r}} in ∑r=0∞cr(1−x)2r+2\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}}
We know that coefficient of xp{{x}^{p}} in (1−x)−m{{\left( 1-x \right)}^{-m}} is (m+p−1)Cp\left( m+p-1 \right){{C}_{p}} or (m+p−1)Cm−1\left( m+p-1 \right){{C}_{m-1}}
Here p is n−1−rn-1-r and m is 2r+22r+2
Coefficient of xn−1−r{{x}^{n-1-r}} in ∑r=0∞cr(1−x)2r+2\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}}=∑r=0∞\sum\limits_{r=0}^{\infty }{{}} (coefficient of xn−1−r{{x}^{n-1-r}} in cr(1−x)2r+2\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}} )

⇒\Rightarrow Coefficient of xn−1−r{{x}^{n-1-r}} in ∑r=0∞cr(1−x)2r+2\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}} = ∑r=0∞\sum\limits_{r=0}^{\infty }{{}} cr×(n−1−r+2r+2−1)C2r+2−1{{c}^{r}}\times \left( n-1-r+2r+2-1 \right){{C}_{2r+2-1}}
⇒\Rightarrow Coefficient of xn−1−r{{x}^{n-1-r}} in ∑r=0∞cr(1−x)2r+2\sum\limits_{r=0}^{\infty }{\dfrac{{{c}^{r}}}{{{\left( 1-x \right)}^{2r+2}}}} = ∑r=0∞(n+r)C2r+1×cr\sum\limits_{r=0}^{\infty }{\left( n+r \right){{C}_{2r+1}}\times {{c}^{r}}}
⇒∑r=0∞(n+r)C2r+1×cr=∑r=0∞(n+r)×(n+r−1)×.........(n−(r−1))×(n−r)×cr2\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 (n+r)×(n−r)\left( n+r \right)\times \left( n-r \right) , (n+r−1)×(n−(r−1))\left( n+r-1 \right)\times \left( n-(r-1) \right) ,……..till (n+1)×(n−1)\left( n+1 \right)\times \left( n-1 \right) and n is left out. We know that (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} so taking n common and multiply these terms we get
⇒∑r=0∞(n+r)×(n+r−1)×.........(n−(r−1))×(n−r)cr∣! 2r+1 ‾=n×∑r=0∞(n2−r2)(n2−(r−1)2).........(n2−22)(n2−12)cr∣! 2r+1 ‾\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
xn{{x}^{n}} in the expansion of x(1−x)2−cx\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≤x≤1-1\le x\le 1
−1≤cx(1−x)2≤1-1\le \dfrac{cx}{{{\left( 1-x \right)}^{2}}}\le 1

Note: Always keep in mind that coefficient of xn{{x}^{n}} in in a series which is in the form xr×S{{x}^{r}}\times S where S is a series will be equal to coefficient of xn−r{{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.