A) \[\frac{{{2}^{n+1}}-1}{n+1}\]
B) \[\frac{{{2}^{n}}-1}{n}\]
C) \[\frac{{{2}^{n-1}}-1}{n-1}\]
D) \[\frac{{{2}^{n+1}}-1}{n+2}\]
Correct Answer: A
Solution :
We know, \[{{(1+x)}^{n}}={{C}_{0}}+{{C}_{1}}x+{{C}_{2}}{{x}^{2}}+....+{{C}_{n}}{{x}^{n}}\] On integrating both sides 0 to 1, we get \[\left[ \frac{{{(1+x)}^{n+1}}}{n+1} \right]_{0}^{1}\] \[=\left[ {{C}_{0}}x+\frac{{{C}_{1}}{{x}^{2}}}{2}+\frac{{{C}_{2}}{{x}^{3}}}{3}+....+\frac{{{C}_{n}}{{x}^{n+1}}}{n+1} \right]_{0}^{1}\] \[\Rightarrow \] \[\frac{{{2}^{n+1}}-1}{n+1}={{C}_{0}}+\frac{{{C}_{1}}}{2}+\frac{{{C}_{2}}}{3}+.....+\frac{{{C}_{n}}}{n+1}\]
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