Question

If $C_0,C_1,C_2...$ are binomial coefficients in the expansion $\sum _{r=0}^n{C}_r{x}^r$ then the value of the expression (series) $\frac{2C_0}{1}+\frac{3C_1}{2}+\frac{4C_2}{3}+\frac{5C_3}{4}+...+$ is

• A. $\frac{2^n+1}{n+1}$
• B. $\frac{2^n-1}{n+1}$
• C. $\frac{2^n(n+3)-1}{n+1}$
• D. $\frac{2^n(n+2)-1}{n+1}$

Answer 1

Answer: (C)

$\frac{2^n(n+3)-1}{n+1}$

Given ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+\cdots +C_n{x}^n$ Integrating both sides with resect to $x$, we get $\frac{{(1+x)}^{n+1}}{n+1}=\frac{C_0{x}^2}{1}+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+\cdots +\frac{C_n{x}^{n+1}}{n+1}+k$ Putting $x=0$ we get $k=\frac{1}{n+1}$ $\therefore \frac{{(1+x)}^{n+1}-1}{n+1}$ $=\frac{C_{0}x}{1}+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+\cdots +\frac{C_n{x}^{n+1}}{n+1}$ multiplying with $x$ both sides $\frac{x{(1+x)}^{n+1}-x}{n+1}=\frac{C_0{x}^2}{1}+\frac{C_1{x}^3}{2}+\frac{C_2{x}^4}{3}+.....+\frac{C_n{x}^{n+2}}{n+1}$ Differentiating with respect to $x$ $\frac{(n+1)\times {(1+x)}^n+{(1+x)}^{n+1}-1}{n+1}$ $\frac{2C_{0}x}{1}+\frac{3C_1{x}^2}{2}+\frac{4C_2{x}^3}{3}+.....+\frac{(n+2)C_n{x}^{n+1}}{n+1}$ Now putting $r=1$ both sides, we get $\frac{2^{n+1}+(n+1)2^n-1}{n+1}$ $=\frac{2C_0}{1}+\frac{3C_1}{2}+\frac{4C_1}{3}+....+\frac{(c+2)C_n}{n+1}$ $\frac{2^n(n+3)-1}{n+1}=\frac{2C_0}{1}+\frac{3C_1}{2}+\frac{4C_2}{3}+\cdots +\frac{(n+2)C_n}{n+1}$