A) \[3{{n}^{2}}+2n+1\]
B) \[2{{n}^{2}}+2n+1\]
C) \[{{n}^{2}}+n+1\]
D) \[2{{n}^{2}}-2n+1\]
Correct Answer: B
Solution :
\[\frac{{{(1+x)}^{2}}}{{{(1-x)}^{3}}}={{(1+x)}^{2}}{{(1-x)}^{-3}}\] \[=(1+2x+{{x}^{2}})[1+{{\,}^{3}}{{C}_{1}}x+{{\,}^{4}}{{C}_{2}}{{x}^{2}}+....\]\[+{{\,}^{.3+n-3}}{{C}_{n-2}}{{x}^{n-2}}+{{\,}^{3+n-2}}{{C}_{n-1}}{{x}^{n-1}}+{{\,}^{3+n-1}}{{C}_{n}}{{x}^{n}}+....]\] \ Coefficient of \[{{x}^{n}}\]\[={{\,}^{3+n-1}}{{C}_{n}}+{{2.}^{3+n-2}}{{C}_{n-1}}{{+}^{3+n-3}}{{C}_{n-2}}\] \[={{\,}^{2+n}}{{C}_{n}}+{{2.}^{n+1}}{{C}_{n-1}}+{{\,}^{n}}{{C}_{n-2}}\] \[=\frac{(2+n)!}{n!2!}+\frac{2.(n+1)!}{(n-1)!2!}+\frac{n!}{(n-2)!2!}\] \[=\frac{n![(2+n)(1+n)+2(n+1)n+n(n-1)]}{n!\,2!}\] \[=2{{n}^{2}}+2n+1\].
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