Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\dots +C_n{x}^n,$ then $\sum _{r=0}^n\sum _{s=0}^n(r+s)C_r{C}_s$ is equal to

• A. $2^{2n}$
• B. $n\cdot 2^{2n}$
• C. $n\cdot 2^{2n}$
• D. None of these

Answer 1

The correct option is C $n\cdot 2^{2n}$

We have,
$\sum _{r=0}^n\sum _{s=0}^n(r+s)C_r{C}_s$
$=\sum _{r=0}^n\sum _{s=0}^n(r\cdot {}^n{C}_r\cdot {}^n{C}_s+s\cdot {}^n{C}_r\cdot {}^n{C}_s)$

$=\sum _{r=0}^n\sum _{s=0}^{n}r\cdot {}^n{C}_r{}^n{C}_s+\sum _{r=0}^n\sum _{s=0}^{n}s\cdot {}^n{C}_r{}^n{C}_s$

$=\sum _{r=0}^{n}r\cdot {}^n{C}_r\left(\sum _{s=0}^n{}^n{C}_s\right)+\sum _{s=0}^{n}s\cdot {}^n{C}_s\left(\sum _{r=0}^n{}^n{C}_r\right)$

$=\sum _{r=0}^{n}r\cdot {}^n{C}_r\cdot 2^n+\sum _{s=0}^{n}s\cdot {}^n{C}_s\cdot 2^n$

$=2^n\left(\sum _{r=0}^{n}r\cdot {}^n{C}_r\right)+2^n\left(\sum _{s=0}^{n}s\cdot {}^n{C}_s\right)$

$=2^n\left(\sum _{r=0}^{n}r\cdot \frac{n}{r}\cdot {}^{n-1}{C}_{r-1}\right)+2^n\left(\sum _{s=0}^{n}s\cdot \frac{n}{s}\cdot {}^{n-1}{C}_{s-1}\right)$

$=2^n(n\cdot 2^{n-1})+2^n\cdot (n\cdot 2^{n-1})$
$=n\cdot 2^{2n-1}+n\cdot 2^{2n-1}$
$=n\cdot 2^{2n}$