A) 6
B) 9
C) 12
D) 24
Correct Answer: C
Solution :
\[{{(1+x)}^{m}}{{(1-x)}^{n}}\] \[=\left( 1+mx+\frac{m(m-1){{x}^{2}}}{2!}+.... \right)\,\left( 1-nx+\frac{n(n-1)}{2!}{{x}^{2}}-.... \right)\] \[=1+(m-n)x+\left[ \frac{{{n}^{2}}-n}{2}-mn+\frac{({{m}^{2}}-m)}{2} \right]{{x}^{2}}\]+......... Given, m - n = 3 or n = m - 3 Hence \[\frac{{{n}^{2}}-n}{2}-mn+\frac{{{m}^{2}}-m}{2}=-6\] Þ \[\frac{(m-3)(m-4)}{2}-m(m-3)+\frac{{{m}^{2}}-m}{2}=-6\] Þ \[{{m}^{2}}-7m+12-2{{m}^{2}}+6m+{{m}^{2}}-m+12=0\] Þ \[-2m+24=0\,\,\,\Rightarrow m=12\]
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