Question

$\frac{{}^{11}{C}_0}{1}$ + $\frac{{}^{11}{C}_1}{2}$ + $\frac{{}^{11}{C}_2}{3}$+ .................... + $\frac{{}^{11}{C}_{1}0}{11}$ =

• A. $\frac{2^{11}-1}{11}$
• B. $\frac{2^{11}-1}{6}$
• C. $\frac{4^{11}-1}{11}$
• D. $\frac{3^{11}-1}{11}$

Answer 1

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

We can use binomial expansion to do this

$(1+x{)}^{11}{=}^{11}{C}_0{+}^{11}{C}_{1}x{+}^{11}{C}_2{x}^2........{.}^{11}{C}_{10}{x}^{10}{+}^{11}{C}_{11}{x}^{11}$

Now integrate the following code

${\int }_0^1(1+x{)}^{11}{=}^{11}{C}_0{+}^{11}{C}_{1}x{+}^{11}{C}_2{x}^2........{.}^{11}{C}_{10}{x}^{10}{+}^{11}{C}_{11}{x}^{11}$

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

Now put $x=1$

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

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

Thus

$\frac{{}^{11}{C}_0}{1}+\frac{{}^{11}{C}_1}{2}+\frac{{}^{11}{C}_2}{3}.........\frac{{}^{11}{C}_{10}}{11}=\frac{2^{12}-2}{12}=\frac{2^{11}-1}{6}$