Question

The sum $\frac{(\frac{11}{0})}{1}+\frac{(\frac{11}{1})}{2}+\frac{(\frac{11}{2})}{3}+.....+\frac{(\frac{11}{11})}{12}$ equals (where $(\frac{n}{r})$ denotes ${}^n{C}_r)$.

• A. $\frac{2^{11}}{12}$
• B. $\frac{2^{12}}{12}$
• C. $\frac{2^{11}-1}{12}$
• D. $\frac{2^{12}-1}{12}$

Answer 1

The correct option is D $\frac{2^{12}-1}{12}$

We have,

${(1+x)}^n=n_{C_0}+n_{C_1}x+n_{C_2}{x}^2+.....+n_{C_n}{x}^n$

Integrating on both sides

${\int }_0^1{(1+x)}^n={\int }_0^1(n_{C_0}+n_{C_1}x+n_{C_2}{x}^2+.....+n_{C_n}{x}^n)dx={\left[\frac{{(1+x)}^{n+1}}{n+1}\right]}_0^1=\frac{n_{C_0}}{1}+\frac{n_{C_1}}{2}+.....+\frac{n_{C_n}}{n+1}=\frac{2^{n+1}}{n+1}-\frac{1}{n+1}=\frac{n_{C_0}}{1}+\frac{n_{C_1}}{2}+.....+\frac{n_{C_n}}{n+1}$

Put $n=11$

$\frac{2^{12}}{12}-\frac{1}{12}=\frac{{11}_{C_0}}{1}+\frac{{11}_{C_1}}{2}+.....+\frac{{11}_{C_n}}{12}⟹\frac{2^{12}-1}{12}=\frac{{11}_{C_0}}{1}+\frac{{11}_{C_1}}{2}+.....+\frac{{11}_{C_n}}{12}$