A) \[\frac{{{2}^{n}}-1}{n+1}\]
B) \[n{{.2}^{n}}\]
C) \[\frac{{{2}^{n}}}{n}\]
D) \[\frac{{{2}^{n}}+1}{n+1}\]
Correct Answer: A
Solution :
We know that \[\frac{{{(1+x)}^{n}}-{{(1-x)}^{n}}}{2}={{C}_{1}}x+{{C}_{3}}{{x}^{3}}+{{C}_{5}}{{x}^{5}}+....\] Integrating from \[x=0\]to \[x=1\], we get \[\frac{1}{2}\int\limits_{0}^{1}{\{{{(1+x)}^{n}}-{{(1-x)}^{n}}\}\,}dx\] \[=\int\limits_{0}^{1}{({{C}_{1}}x+{{C}_{3}}{{x}^{3}}+{{C}_{5}}{{x}^{5}}+....)}dx\] Þ \[\frac{1}{2}\left\{ \frac{{{(1+x)}^{n+1}}}{n+1}+\frac{{{(1-x)}^{n+1}}}{n+1} \right\}_{0}^{1}=\frac{{{C}_{1}}}{2}+\frac{{{C}_{3}}}{4}+\frac{{{C}_{5}}}{6}+....\] or \[\frac{{{C}_{1}}}{2}+\frac{{{C}_{3}}}{4}+\frac{{{C}_{5}}}{6}+....=\frac{1}{2}\left\{ \frac{{{2}^{n+1}}-1}{n+1}+\frac{0-1}{n+1} \right\}\] \[=\frac{1}{2}\left( \frac{{{2}^{n+1}}-2}{n+1} \right)=\frac{{{2}^{n}}-1}{n+1}\]
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