Question

Solve $\frac{{}^{11}{C}_0}{1}+\frac{{}^{11}{C}_1}{2}+\frac{{}^{11}{C}_2}{3}+....+\frac{{}^{11}{C}_{10}}{11}=$

• A. $\frac{2^{11}-1}{11}$
• B. $\frac{2^{11}-1}{6}$
• C. $\frac{3^{11}-1}{11}$
• D. $\frac{3^{11}-1}{6}$

Answer 1

The correct option is B $\frac{2^{11}-1}{6}$

A] We know ${}^{n+1}{C}_{r+1}=\frac{n+1}{r+1}$ ${}^n{C}_r$

$\therefore \sum _{r=0}^{10}\frac{{}^{11}{C}_r}{r+1}$

$=\sum _{r=0}^{10}\frac{{}^{11}{C}_r}{r+1}\frac{12}{12}$

$=\sum _{r=0}^0\frac{{}^{12}{C}_r+1}{12}$

$=\frac{1}{12}{[}^{12}{C}_0{+}^{12}{C}_1+...{+}^{12}{C}_{12}]{-}^{12}{C}_0{-}^{12}{C}_{12}$

$=\frac{1}{12}[2^{12}-2]$

$=\frac{2^{11}-1}{6}$