A) \[\frac{n!}{n!n!}\]
B) \[\frac{(2n)!}{n!n!}\]
C) \[\frac{(2n)!}{n!}\]
D) None of these
Correct Answer: B
Solution :
\[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+.....+{{C}_{n}}{{x}^{n}}\] .....(i) and \[{{\left( 1+\frac{1}{x} \right)}^{n}}={{C}_{0}}+{{C}_{1}}\frac{1}{x}+{{C}_{2}}{{\left( \frac{1}{x} \right)}^{2}}+.....+{{C}_{n}}{{\left( \frac{1}{x} \right)}^{n}}\] ....(ii) If we multiply (i) and (ii), we get \[C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+.....+C_{n}^{2}\] is the term independent of x and hence it is equal to the term independent of x in the product \[{{(1+x)}^{n}}{{\left( 1+\frac{1}{x} \right)}^{n}}\]or in \[\frac{1}{{{x}^{n}}}{{(1+x)}^{2n}}\] or term containing \[{{x}^{n}}\] in \[{{(1+x)}^{2n}}\]. Clearly the coefficient of \[{{x}^{n}}\] in \[{{(1+x)}^{2n}}\]is \[{{T}_{n+1}}\] and equal to \[^{2n}{{C}_{n}}=\frac{(2n)!}{n!\,\,n!}\] Trick : Solving conversely. Put \[n=1,n=2,.....\]then we get \[{{S}_{1}}={{\,}^{1}}C_{0}^{2}+{{\,}^{1}}C_{1}^{2}=2\], \[{{S}_{2}}={{\,}^{2}}C_{0}^{2}+{{\,}^{2}}C_{1}^{2}+{{\,}^{2}}C_{2}^{2}={{1}^{2}}+{{2}^{2}}+{{1}^{2}}=6\] Now check the options (a) Does not hold given condition, (b) (i) Put \[n=1\], then \[\frac{2!}{1!\,1!}=2\] (ii) Put \[n=2\], then \[\frac{4!}{2!\,2!}=\frac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1}=6\] Note: Students should remember this question as an identity.
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