Question

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

• A. $\frac{2^{11}-1}{11}$
• B. $\frac{2^{12}-1}{12}$
• C. $\frac{3^{11}-1}{11}$
• D. $\frac{3^{11}-1}{6}$

Answer 1

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

We know,

$(1+x{)}^{11}{=}^{11}{C}_0{+}^{11}{C}_{1}x{+}^{11}{C}_2{x}^2+....{+}^{11}{C}_{11}{x}^{11}$

Integrating both sides with respect to x:

$\frac{(1+x{)}^{12}}{12}{=}^{11}{C}_0\frac{x}{1}{+}^{11}{C}_1\frac{x^2}{2}{+}^{11}{C}_2\frac{x^3}{3}+....{+}^{11}{C}_{11}\frac{x^{12}}{12}$

Putting x=1

$\frac{(2{)}^{12}}{12}=\frac{{}^{11}{C}_0}{1}+\frac{{}^{11}{C}_1}{2}+\frac{{}^{11}{C}_2}{3}+....+\frac{{}^{11}{C}_{11}}{12}$

$\frac{{}^{11}{C}_0}{1}+\frac{{}^{11}{C}_1}{2}+\frac{{}^{11}{C}_2}{3}+....+\frac{{}^{11}{C}_{10}}{11}=\frac{2^{12}}{12}-\frac{1}{12}=\frac{2^{12}-1}{12}$