Question

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

Answer 1

Consider the problem

It is know that ${\left(r+1\right)}^{th}$ term, $\left(T_{r+1}\right)$, in the binomial expansion of ${\left(a+b\right)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

Assuming that $x^2$ occurs in the ${\left(r+1\right)}^{th}$ term of expansion ${\left(1+x\right)}^m$ we obtain

$T_{r+1}{=}^m{C}_r{\left(1\right)}^{m-r}{\left(x\right)}^r{=}^m{C}_r{\left(x\right)}^r$

Now comparing the indices of $x$ in $x^2$ and in $T_{r+1}$ we obtain

$r=2$

Therefore, the coefficient of $x^2$ is ${}^m{C}_2$

It is given that the coefficient of $x^2$ in the expansion ${\left(1+x\right)}^m$ is $6$

$\begin{array}{c}\therefore {}^m{C}_2=6\\ \Rightarrow \frac{m!}{2!\left(m-2\right)!}=6\\ \Rightarrow \frac{m\left(m-1\right)\left(m-2\right)!}{2\times \left(m-2\right)!}=6\\ \Rightarrow m\left(m-1\right)=12\\ \Rightarrow m^2-m-12=0\\ \Rightarrow m^2-4m+3m-12=0\\ \Rightarrow m\left(m-4\right)+3\left(m-4\right)=0\\ \Rightarrow \left(m-4\right)\left(m-3\right)=0\\ \Rightarrow \left(m-4\right)=0or\left(m+3\right)=0\\ \Rightarrow m=4orm=-3\end{array}$

Thus, the positive value of $m$ for which the coefficient of $x^2$ in the expansion ${\left(1+x\right)}^m$ is $6$ is $4$.