Question

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

• A. $n^2[2^{2n-3}+\frac{1}{2}{}^{2n-2}{C}_{n-1}]$
• B. $[2^{2n-3}-\frac{1}{2}{}^{2n-2}{C}_{n-1}]$
• C. $n^2[2^{2n-3}-\frac{1}{2}{}^{2n-2}{C}_{n-1}]$
• D. $[2^{2n-3}+\frac{1}{2}{}^{2n-2}{C}_{n-1}]$

Answer 1

The correct option is C $n^2[2^{2n-3}-\frac{1}{2}{}^{2n-2}{C}_{n-1}]$

We have, $\underset{0\le r<s\le n}{\sum \sum }(r\cdot s)C_r{C}_s$

$=\underset{0\le r<s\le n}{\sum \sum }(r\cdot {}^n{C}_r)(s\cdot {}^n{C}_s)$

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

$=n^2\cdot \frac{1}{2}[2^{2(n-1)}-{}^{2(n-1)}{C}_{n-1}]$

$=n^2[2^{2n-3}-\frac{1}{2}{}^{2n-2}{C}_{n-1}]$