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, $2.C_0+\frac{2^2}{2}.C_1+\frac{2^3}{2}{C}_2+....+\frac{2^{11}}{11}.C_{10}$ $\sum _{r=0}^{10}{}^{10}{C}_r.\frac{2^{r+1}}{r+1}=\frac{1}{11}\sum _{r=0}^{10}\frac{11}{r+1}.$ ${}^{10}C$ $=\frac{1}{11}\sum _{r=0}^{10}{}^{11}{C}_{r+1}{.2}^{r+1}$ $=\frac{1}{11}{(}^{11}{C}_1{.2}^1+...{+}^{11}{C}_{11}{.2}^{11})$ $=\frac{1}{11}{(}^{11}{C}_0{.2}^0{+}^{11}{C}_1{.2}^1$ $+....{+}^{11}{C}_{11}{.2}^{11}{-}^{11}{C}_0{.2}^0)$ $=\frac{1}{11}[{(1+2)}^{11}-1]=\frac{3^{11}-1}{11}$