Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Each term is of the form $\frac{{}^{11}{C}_r}{r+1}$

${}^{n+1}{C}_{r+1}$ = $\frac{n+1}{r+1}$ ${}^n{C}_r$

⇒ $\frac{{}^n{C}_r}{r+1}$ = $\frac{1}{n+1}$${}^{n+1}{C}_{r+1}$

⇒$\frac{{}^{11}{C}_r}{r+1}$ = $\frac{1}{11+1}$${}^{12}{C}_{r+1}$ = $\frac{1}{12}$${}^{12}{C}_{r+1}$

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

= $\frac{1}{2}$[${}^{12}{C}_1$ + ${}^{12}{C}_2$...................${}^{12}{C}_{11}$]

= $\frac{1}{2}$[${}^{12}{C}_0$ + ${}^{12}{C}_2$...................${}^{12}{C}_{11}$ + ${}^{12}{C}_{12}$ - (${}^{12}{C}_0$ + ${}^{12}{C}_{12}$) ]

(adding and substracting ${}^{12}{C}_0$ + ${}^{12}{C}_{12}$)

= $\frac{1}{12}$ $(2^{12}-2)$

= $\frac{2(2^{11}-1)}{12}$ = $\frac{2^{11}-1}{6}$