Question

Value of $\sum _{r=1}^n\left(\sum _{m=0}^r{}^n{C}_r{.}^r{C}_m\right)$ is equal to :

• A. $2^n-1$
• B. $3^n-1$
• C. $3^n-2^n$
• D. none

Answer 1

The correct option is B $3^n-1$

$\sum _{n=1}^n\left(\sum _{m=0}^r{}^n{C}_r{.}^r{C}_m\right)$

$=\sum _{r=1}^n\left({}^n{C}_r\right(\sum _{m=0}^r{}^r{C}_m\left)\right)$

$=\sum _{r=1}^n{[}^n{C}_r{(}^r{C}_0{+}^r{C}_1{+}^r{C}_2+.............{+}^r{C}_r\left)\right]$

$=\sum _{r=1}^n{[}^n{C}_r{2}^r]$

$[{\because }^n{C}_0{+}^n{C}_1{+}^n{C}_2+...........{+}^n{C}_n=2^n]$

${=}^n{C}_{1}2{+}^n{C}_2{2}^2{+}^n{C}_3{2}^3{+}^n{C}_4{+}^n{C}_5{2}^5+..................{+}^n{C}_n{2}^2$

$=3^n-1$

$[\because (1+x{)}^n{=}^n{C}_0{x}^0{+}^n{C}_1{x}^1{+}^n{C}_2{x}^2+...........{+}^n{C}_n{x}^n]$

at $x=2$

$[\because 3^n=1{+}^n{C}_1(2\left){+}^n{C}_2\right(2{)}^2+........{+}^n{C}_n\left(2{)}^n\right]$

$\Rightarrow sum=3^n-1$