Question

The value of $ 2.{{C}_{0}}+\frac{{{2}^{2}}}{2}.{{C}_{1}} $ $ +\frac{{{2}^{3}}}{3}.{{C}_{2}}+\frac{{{2}^{4}}}{4}.{{C}_{3}}+....+\frac{{{2}^{11}}}{11}.{{C}_{10}} $ is

• A. $ \frac{{{3}^{11}}-1}{11} $
• B. $ \frac{{{2}^{11}}-1}{11} $
• C. $ \frac{{{11}^{3}}-1}{11} $
• D. $ \frac{{{11}^{2}}-1}{11} $

Answer 1

Answer: (A)

We have, $ 2.{{C}_{0}}+\frac{{{2}^{2}}}{2}.{{C}_{1}}+\frac{{{2}^{3}}}{2}{{C}_{2}}+....+\frac{{{2}^{11}}}{11}.{{C}_{10}} $ $ \sum\limits_{r=0}^{10}{^{10}{{C}_{r}}}.\frac{{{2}^{r+1}}}{r+1}=\frac{1}{11}\sum\limits_{r=0}^{10}{\frac{11}{r+1}}. $ $ ^{10}C $ $ =\frac{1}{11}\sum\limits_{r=0}^{10}{^{11}{{C}_{r+1}}{{.2}^{r+1}}} $ $ =\frac{1}{11}{{(}^{11}}{{C}_{1}}{{.2}^{1}}+...{{+}^{11}}{{C}_{11}}{{.2}^{11}}) $ $ =\frac{1}{11}{{(}^{11}}{{C}_{0}}{{.2}^{0}}{{+}^{11}}{{C}_{1}}{{.2}^{1}} $ $ +....{{+}^{11}}{{C}_{11}}{{.2}^{11}}{{-}^{11}}{{C}_{0}}{{.2}^{0}}) $ $ =\frac{1}{11}[{{(1+2)}^{11}}-1]=\frac{{{3}^{11}}-1}{11} $