# Find the positive value of m for which the coefficient of

Question:

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

Solution:

To find: the positive value of m for which the coefficient of x2 in the expansion of $(1+x)^{m}$ is 6

Formula Used:

General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$

Now, finding the general term of the expression, $(1+x)^{m}$, we get

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

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

The coefficient of $(\mathrm{x})^{2}$ is ${ }^{\mathrm{m}} \mathrm{C}_{2}$

${ }^{\mathrm{m}} \mathrm{C}_{2}=6$

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

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

$\mathrm{m}^{2}-\mathrm{m}-6=0$

$(m-3)(m+2)=0$

$m=3,-2$

Since m cannot be negative. Therefore,

m=3

Thus, positive value of $m$ is 3 for which the coefficient of $x 2$ in the expansion of $(1+$ $\mathrm{x})^{\mathrm{m}}$ is 6