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
Show Answer
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,
$m-n=3$
$[\because$ coefficient of $x=3$, given]
and $\quad \frac{1}{2} m(m-1)+\frac{1}{2} n(n-1)-m n=-6$
$$ \begin{aligned} & \Rightarrow \quad m(m-1)+n(n-1)-2 m n=-12 \\ & \Rightarrow \quad m^{2}-m+n^{2}-n-2 m n=-12 \\ & \Rightarrow \quad(m-n)^{2}-(m+n)=-12 \\ & \Rightarrow \quad m+n=9+12=21 \end{aligned} $$
On solving Eqs. (i) and (ii), we get $m=12$
