Question

If $m$ and $n$ are $+ve$ integers and $m>n$ and if $(1+x{)}^{m+n}(1-x{)}^{m-n}$ is expanded as a polynomial in $x$, then the coefficient of $x^2$ is

• A. $2m^2-n$
• B. $2n^2-m$
• C. $2m^2+n$
• D. $2n^2+m$

Answer 1

Answer: (B)

$2n^2-m$

This is a polynomial. Hence we differentiate it twice, the constant of the expression on differentiating twice, would be $2a_2$ where $a_2$ is the coefficient of $x^2$.
$f\left(x\right)=(1+x{)}^{m+n}(1-x)\times m-n$
$f^{\prime }\left(x\right)=(m+n)(1+x{)}^{m+n-1}(1-x)\times m-n-(m-n\left)\right(1+x{)}^{m+n}(1-x)\times m-n-1$
$f^{\prime \prime }\left(x\right)=(m+n)(m+n-1)(1+x{)}^{m+n-2}(1-x)\times m-n-2(m+n\left)\right(m-n\left)\right(1+x{)}^{m+n-1}(1-x)\times m-n-1+(m-n)(m-n-1)(1+x{)}^{m+n}(1-x)\times m-n-2$
Putting $x=0$ to get constant,
Constant $=2a_2=(m+n)(m+n-1)-2(m+n)(m-n)+(m-n)(m-n-1)$

Hence,
$a_2=(m+n)(m+n-1)-2(m+n)(m-n)+(m-n)(m-n-1)=2n^2-m$