question_answer

If\[{{(1+x)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}\,{{x}^{n}},}\]then\[\frac{^{n}{{C}_{0}}}{1.2}{{2}^{2}}+\frac{^{n}{{C}_{1}}}{2.3}{{2}^{3}}+\frac{^{n}{{C}_{2}}}{3.4}{{2}^{4}}+...\]\[+\frac{^{n}{{C}_{2}}}{(n+1).(n+2)}{{2}^{n+2}}\]is equal to

A) \[\frac{{{3}^{n+2}}+2n-5}{(n+1)(n+2)}\]         

B) \[\frac{{{3}^{n+2}}-2n+5}{(n+1)(n+2)}\]

C) \[\frac{{{3}^{n+2}}-2n-5}{(n+1)(n+2)}\]           

D)  None of these

Correct Answer: C

Solution :

Now, \[+\frac{^{n}{{C}_{n}}}{(n+1).(n+2)}{{2}^{n+2}}\] \[=\sum\limits_{r=0}^{n}{{}}\frac{1}{(r+1)(r+2)}{{\,}^{n}}{{C}_{r}}{{.2}^{r+2}}\] \[=\frac{1}{(n+1)(n+2)}\sum\limits_{r=0}^{n}{\frac{n+2}{r+2}\frac{n+1}{r+1}}{{\,}^{n}}{{C}_{r}}{{2}^{r+2}}\]\[=\frac{1}{(n+1)(n+2)}\sum\limits_{r=0}^{n}{^{n+2}}{{C}_{r+2}}{{2}^{r+2}}\] \[=\frac{1}{(n+1)(n+2)}\left\{ \sum\limits_{s=0}^{n}{^{n+2}}{{C}_{s}}{{2}^{s}} \right\}\] \[=\frac{1}{(n+1)(n+2)}\left[ \left\{ \sum\limits_{s=0}^{n+2}{^{n+2}}{{C}_{s}}{{2}^{s}} \right\} \right.\] \[-\left. \left\{ ^{n+2}{{C}_{0}}{{2}^{0}}{{+}^{n+21}}{{C}_{1}}{{2}^{1}} \right\} \right]\] \[=\frac{1}{(n+1)(n+2)}{{[1+2)}^{n+2}}-\{1+2(n+2)\}]\] \[=\frac{{{3}^{n+2}}-2n-5}{(n+1)(n+2)}\]