A) \[\frac{1}{n+1}\]
B) \[\frac{1}{n+2}\]
C) \[\frac{1}{n(n+1)}\]
D) None of these
Correct Answer: D
Solution :
\[{{(1-x)}^{n}}={{C}_{0}}-{{C}_{1}}x+{{C}_{2}}{{x}^{2}}-{{C}_{3}}{{x}^{3}}+.....\] Þ \[x\,\,{{(1-x)}^{n}}={{C}_{0}}x-{{C}_{1}}{{x}^{2}}+{{C}_{2}}{{x}^{3}}-{{C}_{3}}{{x}^{4}}+.....\] Þ \[\int\limits_{0}^{1}{x}{{(1-x)}^{n}}dx=\int\limits_{0}^{1}{({{C}_{0}}x-{{C}_{1}}{{x}^{2}}+{{C}_{2}}{{x}^{3}}....)dx}\] ?...(i) The integral on the LHS \[=\int\limits_{1}^{0}{(1-t){{t}^{n}}(-dt),}\]by putting \[1-x=t\] \[=\int\limits_{0}^{1}{({{t}^{n}}-{{t}^{n+1}})}\,dt=\frac{1}{n+1}-\frac{1}{n+2}\] Whereas the integral on the RHS of (i) \[=\left[ \frac{{{C}_{0}}{{x}^{2}}}{2}-\frac{{{C}_{1}}{{x}^{3}}}{3}+\frac{{{C}_{2}}{{x}^{4}}}{4}-.... \right]\]\[=\frac{{{C}_{0}}}{2}-\frac{{{C}_{1}}}{3}+\frac{{{C}_{2}}}{4}-....\] \[\therefore \,\,\,\,\frac{{{C}_{0}}}{2}-\frac{{{C}_{1}}}{3}+\frac{{{C}_{2}}}{4}-....\]to \[(n+1)\] terms \[=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}\].
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