Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+\dots +C_n{x}^n,$ then
the value of $\underset{0\le r<s\le n}{\sum \sum }(r+s)(C_r+C_s)$ is

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

Answer 1

The correct option is A $n^2\cdot 2^n$

We have,
$\underset{0\le r,s\le n}{\sum \sum }(r+s)(C_r+C_s)$
$=\sum _{r=0}^n\sum _{s=0}^n(r{C}_r+r{C}_s+s{C}_r+s{C}_s)$
$=\sum _{r=0}^n[\sum _{s=0}^{n}r{C}_r+\sum _{s=0}^{n}r{C}_s+\sum _{s=0}^{n}s{C}_r+\sum _{s=0}^{n}s{C}_s]$
$=\sum _{r=0}^n\left[\right(n+1)r{C}_r+r{2}^n+\frac{n(n+1)}{2}{C}_r+n\cdot 2^{n-1}]$
$=(n+1)(n\cdot 2^{n-1})+2^n\frac{n(n+1)}{2}+\frac{n(n+1)}{2}{2}^n+n\cdot 2^{n-1}(n+1)$
$=n(n+1)2^n+n(n+1)2^n$
$=2n(n+1)2^n$

Also
$\underset{0\le r,s\le n}{\sum \sum }(r+s)(C_r+C_s)$
$=\sum _{r=0}^{n}4r{C}_r+2\underset{0\le r<s\le n}{\sum \sum }(r+s)(C_r+C_s)$
$\therefore 2n(n+1)2^n=4n\cdot 2^{n-1}+2\underset{0\le r<s\le n}{\sum \sum }(r+s)(C_r+C_s)$
$\Rightarrow \underset{0\le r<s\le n}{\sum \sum }(r+s)(C_r+C_s)=n^2\cdot 2^n$