A) \[\frac{1}{n(n+1)}\]
B) \[\frac{1}{(n+1)(n+2)}\]
C) \[\frac{1}{(n+1)(n+3)}\]
D) \[\frac{1}{n(n+3)}\]
E) \[\frac{1}{(n+2)(n+3)}\]
Correct Answer: B
Solution :
\[{{(1-x)}^{n}}={{C}_{0}}-{{C}_{1}}x+{{C}_{2}}{{x}^{2}}-{{C}_{3}}{{x}^{3}}\] \[+.....+{{(-1)}^{n}}{{C}_{n}}.{{x}^{n}}\] ?.(i) \[\Rightarrow \]\[x{{(1-x)}^{n}}=x{{C}_{0}}-{{C}_{1}}{{x}^{2}}+{{C}_{2}}{{x}^{3}}-{{C}_{3}}{{x}^{4}}\] \[+.....+{{(-1)}^{n}}{{C}_{n}}{{x}^{n+1}}\] On integrating w.r.t.\[x:\]between 0 to 1. \[\left[ -x\frac{{{(1-x)}^{n+1}}}{n+1}-\frac{{{(1-x)}^{n+2}}}{(n+1)(n+2)} \right]_{0}^{1}\] \[=\left[ \frac{{{x}^{2}}{{C}_{0}}}{2}-\frac{{{C}_{1}}{{x}^{3}}}{3}+\frac{{{C}_{2}}{{x}^{4}}}{4}.....{{(-1)}^{n}}\frac{{{C}_{n}}{{x}^{n+2}}}{n+2} \right]_{0}^{1}\] \[\Rightarrow \] \[\frac{{{(1-0)}^{n+2}}}{(n+1)(n+2)}=\frac{{{C}_{0}}}{2}-\frac{{{C}_{1}}}{3}+\frac{{{C}_{2}}}{4}\] \[+...{{(-1)}^{n}}\frac{{{C}_{n}}}{n+2}\] \[\Rightarrow \] \[\frac{{{C}_{0}}}{2}-\frac{{{C}_{1}}}{3}+\frac{{{C}_{2}}}{4}-\frac{{{C}_{3}}}{5}+...+{{(-1)}^{n}}\frac{{{C}_{n}}}{n+2}\] \[=\frac{1}{(n+1)(n+2)}\]
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