Question

$\stackrel{50}{\sum _{r=0}}\frac{(r+2)}{(r+1)}{(}^{50}{C}_r)$ is-

• A. $2^{50}+\frac{(2^{51}-1)}{51}$
• B. $\frac{(2^{50}-1)}{51}$
• C. $2^{51}+\frac{(2^{150}-1)}{51}$
• D. $2^{50}+\frac{(2^{150}-1)}{52}$

Answer 1

The correct option is A $2^{50}+\frac{(2^{51}-1)}{51}$

$\sum _{r=0}^{50}50cr+\sum _{r=0}^{50}\frac{50cr}{r+1}$

$\underset{2^{50}}{\underset{}{\sum _{r=0}^{50}50cr}}+\sum _{r=0}^{50}\frac{50cr}{r+1}$ $\downarrow \frac{50cr}{1}+\frac{50cr}{2}$

$(1+x{)}^n=1{+}^n{C}_{1}x{+}^n{C}_2{x}^2{+}^n{C}_3{x}^3$...........

Integrating from $0$ to $1$ on both sides.

${\begin{array}{c}\frac{(1+x{)}^{n+1}}{n+1}\end{array}|}_0^1$ ${\begin{array}{c}\frac{x{+}^n{C}_1{x}^2}{2}+\frac{{}^n{C}_2{x}^3}{3}\end{array}|}_0^1$

$\frac{2^{n+1}}{n+1}-\frac{1}{n+1}=1+\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{3}$

$n=50$

$\frac{2^{51}}{51}-\frac{1}{51}=50C_0+\frac{50C_1}{2}$

$2^{50}+\frac{2^{51}-1}{51}\to$ Requaired sum