A) \[\frac{(2n)!}{(n-r)\,!\,(n+r)!}\]
B) \[\frac{n!}{(-r)!(n+r)!}\]
C) \[\frac{n!}{(n-r)!}\]
D) None of these
Correct Answer: A
Solution :
\[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+....+{{C}_{r}}{{x}^{r}}+....\] ?..(i) \[{{\left( 1+\frac{1}{x} \right)}^{n}}={{C}_{0}}+{{C}_{1}}\frac{1}{x}+{{C}_{2}}\frac{1}{{{x}^{2}}}+.....+{{C}_{r}}\frac{1}{{{x}^{r}}}+....\] ?..(ii) Multiplying both sides and equating coefficient of \[{{x}^{r}}\]in \[\frac{1}{{{x}^{n}}}{{(1+x)}^{2n}}\]or the coefficient of \[{{x}^{n+r}}\]in \[{{(1+x)}^{2n}}\] we get the value of required expression = \[^{2n}{{C}_{n+r}}=\frac{(2n)\,!}{(n-r)\,!\,(n+r)\,!}\] Trick: Solving conversely. Put \[n=1\] and \[r=0\] in first term , (given condition) (i) \[^{1}{{C}_{0}}^{1}{{C}_{0}}{{+}^{1}}{{C}_{1}}^{1}{{C}_{1}}=1+1=2\] , \[(\because r\le n)\] Put \[n=2,r=1\], then (ii) \[^{2}{{C}_{0}}^{2}{{C}_{1}}{{+}^{2}}{{C}_{1}}^{2}{{C}_{2}}=2+2=4\] Now check the options (i) Put \[n=1,r=0\], we get \[\frac{2!}{(1)!\,\,(1)!}=2\] (ii) Put\[n=2,r=1\], we get \[\frac{4!}{(1)\,!\,\,(3)\,!}=4\] Note: Students should remember this question as an identity.
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