Question

Find a positive value of $m$ for which the coefficient of $x^2$ in the expansion ${(1+x)}^m$ is $6$

Answer 1

We know that

General term of expansion ${(a+b)}^n$ is $T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

General term of expansion ${(1+x)}^m$ is

$T_{r+1}{=}^m{C}_r{1}^{m-r}{x}^r{=}^m{C}_r{x}^r$

Thus,$T_{r+1}{=}^m{C}_r{x}^r$

We need coefficient of $x^2$

So,$T_2{=}^m{C}_2{x}^2$

$\therefore$ coefficient of $x^2$ is ${}^m{C}_{2}6$

$\Rightarrow \frac{m!}{2!(m-2)!}=6$

$\Rightarrow \frac{m(m-1)(m-2)!}{2(m-2)!}=6$

$\Rightarrow \frac{m(m-1)}{2}=6$

$\Rightarrow m(m-1)=12$

$\Rightarrow m^2-m-12=0$

$\Rightarrow m^2-4m+3m-12=0$

$\Rightarrow m(m-4)+3(m-4)=0$

$\Rightarrow (m-4)(m+3)=0$

$\therefore m=4,-3$

But we need positive value of $m$

Hence $m=4$