Question

If coefficient of $x^2$ in the expansion of $(1+x{)}^m$ is $6$. Then $m=$ ?

• A. $4$
• B. $-4$
• C. $0$
• D. $none$

Answer 1

The correct option is A $4$

As we know the binomial expansion-

${(1+x)}^m{=}^m{c}_0{1}^m{+}^m{c}_1{1}^{m-1}\cdot x{+}^m{c}_2{1}^{m-2}\cdot x^2+......$

Comparing the coefficient of $x^2$ with $6$, we get

${}^m{c}_2=6$

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

$m(m-1)=12$

$m^2-m-12=0$

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

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

$m=4,-3\left(Notconsiderable\right)$

Hence, $m=4.$