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

Open in App

Solution

Here coefficient of $x^{2}$ in $(1 + x)^{m} = m_{C_{2}} = 6$

$∴ \frac{m (m - 1)}{2} = 6$

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

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

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

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

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

Either m-4=0 or m+3=0

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

So positive value of m is 4.

Suggest Corrections

Similar questions

m

x^{2}

{(1 + x)}^{m}

6

x^{2}

(1 + x)^{m}

6

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

x^{2}

(1 + x)^{m}

6

x^{2}

(1 + x)^{m}

6

m =

Join BYJU'S Learning Program