Binomial Theorem 1 Question 21
23.
If in the expansion of $(1+x)^{m}(1-x)^{n}$, the coefficients of $x$ and $x^{2}$ are $3$ and $-6$ respectively, then $m$ is euqal to
(1999, 2M)
(a) $6$
(b) $9$
(c) $12$
(d) $24$
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Answer:
Correct Answer: 23. (c)
Solution:
- $(1+x)^{m}(1-x)^{n}=[1+m x+\frac{m(m-1)}{2} x^{2}+\ldots]$
$ [1-n x+\frac{n(n-1)}{2} x^{2}-\ldots] $
$ =1+(m-n) x+[\frac{m(m-1)}{2}+\frac{n(n-1)}{2}-m n ]x^{2}+\ldots $
term containing power of $x \geq 3$.
Now, $m-n=3$ $\quad$ …….(i)
$[\because$ coefficient of $x=3$, given]
and $\quad \frac{1}{2} m(m-1)+\frac{1}{2} n(n-1)-m n=-6$
$ \begin{aligned} & \Rightarrow \quad m(m-1)+n(n-1)-2 m n=-12 \\ & \Rightarrow \quad m^{2}-m+n^{2}-n-2 m n=-12 \\ & \Rightarrow \quad(m-n)^{2}-(m+n)=-12 \\ & \Rightarrow \quad m+n=9+12=21 \quad …….(ii) \end{aligned} $
On solving Eqs. (i) and (ii), we get $m=12$
