Find a positive value of m for which the coefficient of x2 in the expansion

Solution:

It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right)$, in the binomial expansion of $(a+b)^{n}$ is given by $\mathrm{T}_{\mathrm{r}+1}={ }^{n} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{t}}$.

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

$\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{m}} \mathrm{C}_{\mathrm{r}}(1)^{\mathrm{m}-\mathrm{t}}(\mathrm{x})^{\mathrm{r}}={ }^{\mathrm{m}} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{\mathrm{t}}$

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

$r=2$

Therefore, the coefficient of $x^{2}$ is ${ }^{\mathrm{m}} \mathrm{C}_{2}$.

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

$\therefore^{m} C_{2}=6$

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

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

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

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

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

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

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

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

$\Rightarrow m=4$ or $m=-3$

Thus, the positive value of $m$, for which the coefficient of $x^{2}$ in the expansion

$(1+x)^{m}$ is 6, is 4