Question

If ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ are the respective coefficient of $x^m$ and $x^n$ in the expansion of $(1+x{)}^{m+n}$, then

• A. $a+b=m+n$
• B. $a=2b$
• C. $a=b$
• D. $b=2a$

Answer 1

The correct option is C $a=b$

Using Binomial theorem,

$(1+x{)}^{m+n}=\sum _{k=0}^{m+n}\left(\begin{array}{c}m+n\\ k\end{array}\right)x^k$

$T_{k+1}=\left(\begin{array}{c}m+n\\ k\end{array}\right)x^k$

$\therefore$ Coefficient of $x^m,a=\left(\begin{array}{c}m+n\\ m\end{array}\right)=\frac{(m+n)!}{m!(m+n-m)!}$

$a=\frac{(m+n)!}{m!n!}$

Coefficient of $x^n,b=\left(\begin{array}{c}m+n\\ n\end{array}\right)=\frac{(m+n)!}{n!(m+n-n)!}$

$b=\frac{m+n)!}{n!m!}$

$\therefore a+b=\frac{(m+n)!}{m!n!}+\frac{(m+n)!}{m!n!}$

$\Rightarrow a+b=2a$

$\Rightarrow b=2a-a$

$\Rightarrow b=a$.