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Q12. Find a positive value of m for which the coefficient of $x^2$ in the expansion $(1 + x)^ m$ is 6.

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As we know that the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(a+b)^n$ is given by

$T_{r+1}=^nC_ra^{n-r}b^r$

So, the general $(r+1)^{th}$ term $T_{r+1}$ in the binomial expansion of $(1 + x)^ m$ is

$T_{r+1}=^mC_r1^{m-r}x^r=^mC_rx^r$

$x^2$ will come when $r=2$ . So,

The coefficient of $x^2$ in the binomial expansion of $(1 + x)^ m$ = 6

$\Rightarrow ^mC_2=6$

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

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

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

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

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

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

Hence the positive value of m for which the coefficient of $x^2$ in the expansion $(1 + x)^ m$ is 6, is 4.

Posted by

Pankaj Sanodiya

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