Question

$\frac{C_0}{1}+\frac{C_1}{2}+\frac{C_2}{3}+......+\frac{C_{10}}{11}=$

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

Answer 1

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

We know that $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+...+C_n{x}^n$

Integrate both sides from $0$ to $1$ we get

${\int }_0^1(1+x{)}^{n}dx={\int }_0^1[C_0+C_{1}x+C_2{x}^2+...+C_n{x}^{n+1}]dx$

$\Rightarrow {\left[\frac{(1+x{)}^{n+1}}{n+1}\right]}_0^1={[C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+...+C_n\frac{x^{n+1}}{n+1}]}_0^1$

$\Rightarrow C_0+\frac{C_1}{2}+\frac{C_2}{3}+...+\frac{C_n}{n+1}=\frac{2^{n+1}-1}{(n+1)}...\left(1\right)$

Put $n=10$ in $\left(1\right)$ we get

$\frac{C_0}{1}+\frac{C_1}{2}+\frac{C_2}{3}+...+\frac{C_{10}}{11}=\frac{2^{11}-1}{11}$