Find a, b, c so that the coefficient of ${{x}^{n}}$ in the expansion of $\dfrac{a+bx+c{{x}^{2}}}{{{\left( 1-x \right)}^{3}}}$ may be ${{n}^{2}}+1$.
Answer
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Hint: First of all write the expression given above $\dfrac{a+bx+c{{x}^{2}}}{{{\left( 1-x \right)}^{3}}}$ as $\left( a+bx+c{{x}^{2}} \right){{\left( 1-x \right)}^{-3}}$. We know that the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1-x \right)}^{-n}}$ is ${}^{n+r-1}{{C}_{r}}$ so using this formula in finding the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1-x \right)}^{-3}}$ we get, ${}^{3+r-1}{{C}_{r}}$ or ${}^{2+r}{{C}_{r}}$. Now, find the coefficient of ${{x}^{n}}$ in $a{{\left( 1-x \right)}^{-3}}+bx{{\left( 1-x \right)}^{-3}}+c{{x}^{2}}{{\left( 1-x \right)}^{-3}}$ by using different value of r in the coefficient of ${{x}^{r}}$ in such a way that we get the power of x as n. After finding the coefficient of ${{x}^{n}}$ equate it to ${{n}^{2}}+1$ and then compare both the sides of the equation to get the value of a, b and c.
Complete step-by-step solution -
The expression given in the question of which expansion we have to found is:
$\dfrac{a+bx+c{{x}^{2}}}{{{\left( 1-x \right)}^{3}}}$
Rewriting the above expression we get,
$\begin{align}
& \left( a+bx+c{{x}^{2}} \right){{\left( 1-x \right)}^{-3}} \\
& \Rightarrow a{{\left( 1-x \right)}^{-3}}+bx{{\left( 1-x \right)}^{-3}}+c{{x}^{2}}{{\left( 1-x \right)}^{-3}} \\
\end{align}$
We know that the formula for the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1-x \right)}^{-n}}$ is:
${}^{n+r-1}{{C}_{r}}$
Using the above formula we can find the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1-x \right)}^{-3}}$ as follows:
$\begin{align}
& {}^{3+r-1}{{C}_{r}} \\
& ={}^{2+r}{{C}_{r}} \\
\end{align}$
Now, we are going to find the coefficient of ${{x}^{n}}$ in $a{{\left( 1-x \right)}^{-3}}+bx{{\left( 1-x \right)}^{-3}}+c{{x}^{2}}{{\left( 1-x \right)}^{-3}}$ using the above formula we get,
We are breaking the expression $a{{\left( 1-x \right)}^{-3}}+bx{{\left( 1-x \right)}^{-3}}+c{{x}^{2}}{{\left( 1-x \right)}^{-3}}$ in three parts as follows:
$a{{\left( 1-x \right)}^{-3}},bx{{\left( 1-x \right)}^{-3}},c{{x}^{2}}{{\left( 1-x \right)}^{-3}}$
Then finding the coefficient of ${{x}^{n}}$ in each of the above expressions and then adding them.
The coefficient of ${{x}^{n}}$ in $a{{\left( 1-x \right)}^{-3}}$ is:
$a\left( {}^{2+n}{{C}_{n}} \right)$ …………… Eq. (1)
The coefficient of ${{x}^{n}}$ in $bx{{\left( 1-x \right)}^{-3}}$ is:
In the expression $bx{{\left( 1-x \right)}^{-3}}$ we have already had x so we have to find the coefficient of ${{x}^{n-1}}$ in ${{\left( 1-x \right)}^{-3}}$ to get the coefficient of ${{x}^{n}}$ which is:
$\begin{align}
& b\left( {}^{2+r}{{C}_{r}} \right) \\
& =b\left( {}^{2+n-1}{{C}_{n-1}} \right) \\
& =b\left( {}^{n+1}{{C}_{n-1}} \right).............Eq.(2) \\
\end{align}$
As $c{{x}^{2}}{{\left( 1-x \right)}^{-3}}$ contains ${{x}^{2}}$ so to get the coefficient of ${{x}^{n}}$ we need to find the coefficient of ${{x}^{n-2}}$ in ${{\left( 1-x \right)}^{-3}}$ which we have shown below:
$c\left( {}^{2+r}{{C}_{r}} \right)$
Substituting r as n – 2 in the above expression we get,
$\begin{align}
& c\left( {}^{2+n-2}{{C}_{n-2}} \right) \\
& =c\left( {}^{n}{{C}_{n-2}} \right)...........Eq.(3) \\
\end{align}$
Adding eq. (1, 2 & 3) we get,
$a\left( {}^{2+n}{{C}_{n}} \right)+b\left( {}^{n+1}{{C}_{n-1}} \right)+c\left( {}^{n}{{C}_{n-2}} \right)$
Using the below relation ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ in the above expression we get,
$a\left( {}^{n+2}{{C}_{2}} \right)+b\left( {}^{n+1}{{C}_{2}} \right)+c\left( {}^{n}{{C}_{2}} \right)$
We know that:
${}^{n}{{C}_{r}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)......\left( n-\left( r-1 \right) \right)}{r!}$
Using this relation in $a\left( {}^{n+2}{{C}_{2}} \right)+b\left( {}^{n+1}{{C}_{2}} \right)+c\left( {}^{n}{{C}_{2}} \right)$ we get,
$\begin{align}
& a\left( \dfrac{\left( n+2 \right)\left( n+1 \right)}{2!} \right)+b\left( \dfrac{\left( n+1 \right)\left( n \right)}{2!} \right)+c\left( \dfrac{n\left( n-1 \right)}{2!} \right) \\
& =\dfrac{a\left( n+2 \right)\left( n+1 \right)+b\left( n+1 \right)\left( n \right)+cn\left( n-1 \right)}{2.1} \\
& =\dfrac{a\left( {{n}^{2}}+3n+2 \right)+b\left( {{n}^{2}}+n \right)+c{{n}^{2}}-cn}{2} \\
\end{align}$
Separating the coefficient of ${{n}^{2}}$ and the coefficient of n in the above expression we get,
$\begin{align}
& \dfrac{{{n}^{2}}\left( a+b+c \right)+n\left( 3a+b-c \right)+2a}{2} \\
& =\dfrac{{{n}^{2}}\left( a+b+c \right)+n\left( 3a+b-c \right)}{2}+a \\
\end{align}$
Now, it is given that the coefficient of ${{x}^{n}}$ in the expansion of $\dfrac{a+bx+c{{x}^{2}}}{{{\left( 1-x \right)}^{3}}}$ is equal to ${{n}^{2}}+1$ so equating the above expression to ${{n}^{2}}+1$ we get,
$\dfrac{{{n}^{2}}\left( a+b+c \right)+n\left( 3a+b-c \right)}{2}+a={{n}^{2}}+1$
Comparing the coefficient of ${{n}^{2}},n$ and the terms which are independent of n on both the sides we get,
$\begin{align}
& \dfrac{a+b+c}{2}=1......Eq.(4) \\
& \dfrac{3a+b-c}{2}=0.......Eq.(5) \\
& a=1.......Eq.(6) \\
\end{align}$
From eq. (6) we have got the value of “a” as 1. Substituting the value of “a” in eq. (4) and eq. (5) we get,
$\dfrac{1+b+c}{2}=1$
Cross – multiplying both the sides we get,
$\begin{align}
& 1+b+c=2 \\
& \Rightarrow b+c=1........Eq.(7) \\
\end{align}$
$\dfrac{3\left( 1 \right)+b-c}{2}=0$
Cross – multiplying on both the sides of the above equation we get,
$\begin{align}
& 3+b-c=0 \\
& \Rightarrow b-c=-3......Eq.(8) \\
\end{align}$
Adding eq. (7) and eq. (8) we get,
$\begin{align}
& \text{ }b+c=1 \\
& \dfrac{+b-c=-3}{2b=-2} \\
\end{align}$
Dividing 2 on both the sides we get,
$b=-1$
Substituting the above value of b in eq. (7) we get,
$\begin{align}
& b+c=1 \\
& \Rightarrow -1+c=1 \\
& \Rightarrow c=2 \\
\end{align}$
From the above solution, we have got the value of a, b and c as 1, -1 and 2 respectively.
Note: In the first reading of the question, you might have thought that the expression is not containing any value of ${{x}^{n}}$ so from where this coefficient of ${{x}^{n}}$ will come. The whole story lies in the expansion of ${{\left( 1-x \right)}^{-3}}$ if you know this binomial expansion then you can easily solve the question so you must know the expansion of ${{\left( 1-x \right)}^{-n}}$.
Complete step-by-step solution -
The expression given in the question of which expansion we have to found is:
$\dfrac{a+bx+c{{x}^{2}}}{{{\left( 1-x \right)}^{3}}}$
Rewriting the above expression we get,
$\begin{align}
& \left( a+bx+c{{x}^{2}} \right){{\left( 1-x \right)}^{-3}} \\
& \Rightarrow a{{\left( 1-x \right)}^{-3}}+bx{{\left( 1-x \right)}^{-3}}+c{{x}^{2}}{{\left( 1-x \right)}^{-3}} \\
\end{align}$
We know that the formula for the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1-x \right)}^{-n}}$ is:
${}^{n+r-1}{{C}_{r}}$
Using the above formula we can find the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1-x \right)}^{-3}}$ as follows:
$\begin{align}
& {}^{3+r-1}{{C}_{r}} \\
& ={}^{2+r}{{C}_{r}} \\
\end{align}$
Now, we are going to find the coefficient of ${{x}^{n}}$ in $a{{\left( 1-x \right)}^{-3}}+bx{{\left( 1-x \right)}^{-3}}+c{{x}^{2}}{{\left( 1-x \right)}^{-3}}$ using the above formula we get,
We are breaking the expression $a{{\left( 1-x \right)}^{-3}}+bx{{\left( 1-x \right)}^{-3}}+c{{x}^{2}}{{\left( 1-x \right)}^{-3}}$ in three parts as follows:
$a{{\left( 1-x \right)}^{-3}},bx{{\left( 1-x \right)}^{-3}},c{{x}^{2}}{{\left( 1-x \right)}^{-3}}$
Then finding the coefficient of ${{x}^{n}}$ in each of the above expressions and then adding them.
The coefficient of ${{x}^{n}}$ in $a{{\left( 1-x \right)}^{-3}}$ is:
$a\left( {}^{2+n}{{C}_{n}} \right)$ …………… Eq. (1)
The coefficient of ${{x}^{n}}$ in $bx{{\left( 1-x \right)}^{-3}}$ is:
In the expression $bx{{\left( 1-x \right)}^{-3}}$ we have already had x so we have to find the coefficient of ${{x}^{n-1}}$ in ${{\left( 1-x \right)}^{-3}}$ to get the coefficient of ${{x}^{n}}$ which is:
$\begin{align}
& b\left( {}^{2+r}{{C}_{r}} \right) \\
& =b\left( {}^{2+n-1}{{C}_{n-1}} \right) \\
& =b\left( {}^{n+1}{{C}_{n-1}} \right).............Eq.(2) \\
\end{align}$
As $c{{x}^{2}}{{\left( 1-x \right)}^{-3}}$ contains ${{x}^{2}}$ so to get the coefficient of ${{x}^{n}}$ we need to find the coefficient of ${{x}^{n-2}}$ in ${{\left( 1-x \right)}^{-3}}$ which we have shown below:
$c\left( {}^{2+r}{{C}_{r}} \right)$
Substituting r as n – 2 in the above expression we get,
$\begin{align}
& c\left( {}^{2+n-2}{{C}_{n-2}} \right) \\
& =c\left( {}^{n}{{C}_{n-2}} \right)...........Eq.(3) \\
\end{align}$
Adding eq. (1, 2 & 3) we get,
$a\left( {}^{2+n}{{C}_{n}} \right)+b\left( {}^{n+1}{{C}_{n-1}} \right)+c\left( {}^{n}{{C}_{n-2}} \right)$
Using the below relation ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ in the above expression we get,
$a\left( {}^{n+2}{{C}_{2}} \right)+b\left( {}^{n+1}{{C}_{2}} \right)+c\left( {}^{n}{{C}_{2}} \right)$
We know that:
${}^{n}{{C}_{r}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)......\left( n-\left( r-1 \right) \right)}{r!}$
Using this relation in $a\left( {}^{n+2}{{C}_{2}} \right)+b\left( {}^{n+1}{{C}_{2}} \right)+c\left( {}^{n}{{C}_{2}} \right)$ we get,
$\begin{align}
& a\left( \dfrac{\left( n+2 \right)\left( n+1 \right)}{2!} \right)+b\left( \dfrac{\left( n+1 \right)\left( n \right)}{2!} \right)+c\left( \dfrac{n\left( n-1 \right)}{2!} \right) \\
& =\dfrac{a\left( n+2 \right)\left( n+1 \right)+b\left( n+1 \right)\left( n \right)+cn\left( n-1 \right)}{2.1} \\
& =\dfrac{a\left( {{n}^{2}}+3n+2 \right)+b\left( {{n}^{2}}+n \right)+c{{n}^{2}}-cn}{2} \\
\end{align}$
Separating the coefficient of ${{n}^{2}}$ and the coefficient of n in the above expression we get,
$\begin{align}
& \dfrac{{{n}^{2}}\left( a+b+c \right)+n\left( 3a+b-c \right)+2a}{2} \\
& =\dfrac{{{n}^{2}}\left( a+b+c \right)+n\left( 3a+b-c \right)}{2}+a \\
\end{align}$
Now, it is given that the coefficient of ${{x}^{n}}$ in the expansion of $\dfrac{a+bx+c{{x}^{2}}}{{{\left( 1-x \right)}^{3}}}$ is equal to ${{n}^{2}}+1$ so equating the above expression to ${{n}^{2}}+1$ we get,
$\dfrac{{{n}^{2}}\left( a+b+c \right)+n\left( 3a+b-c \right)}{2}+a={{n}^{2}}+1$
Comparing the coefficient of ${{n}^{2}},n$ and the terms which are independent of n on both the sides we get,
$\begin{align}
& \dfrac{a+b+c}{2}=1......Eq.(4) \\
& \dfrac{3a+b-c}{2}=0.......Eq.(5) \\
& a=1.......Eq.(6) \\
\end{align}$
From eq. (6) we have got the value of “a” as 1. Substituting the value of “a” in eq. (4) and eq. (5) we get,
$\dfrac{1+b+c}{2}=1$
Cross – multiplying both the sides we get,
$\begin{align}
& 1+b+c=2 \\
& \Rightarrow b+c=1........Eq.(7) \\
\end{align}$
$\dfrac{3\left( 1 \right)+b-c}{2}=0$
Cross – multiplying on both the sides of the above equation we get,
$\begin{align}
& 3+b-c=0 \\
& \Rightarrow b-c=-3......Eq.(8) \\
\end{align}$
Adding eq. (7) and eq. (8) we get,
$\begin{align}
& \text{ }b+c=1 \\
& \dfrac{+b-c=-3}{2b=-2} \\
\end{align}$
Dividing 2 on both the sides we get,
$b=-1$
Substituting the above value of b in eq. (7) we get,
$\begin{align}
& b+c=1 \\
& \Rightarrow -1+c=1 \\
& \Rightarrow c=2 \\
\end{align}$
From the above solution, we have got the value of a, b and c as 1, -1 and 2 respectively.
Note: In the first reading of the question, you might have thought that the expression is not containing any value of ${{x}^{n}}$ so from where this coefficient of ${{x}^{n}}$ will come. The whole story lies in the expansion of ${{\left( 1-x \right)}^{-3}}$ if you know this binomial expansion then you can easily solve the question so you must know the expansion of ${{\left( 1-x \right)}^{-n}}$.
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