Question

The sum ${\sum }_{i=0}^m(\frac{10}{i})(\frac{20}{m-i})$ where $(\frac{p}{q})=0$; If (p<q) is maximum when m is

• A. 5
• B. 10
• C. 15
• D. 20

Answer 1

The correct option is C 15

To find ${\sum }_{i=0}^m(\frac{10}{i})(\frac{20}{m-i})$

${\sum }_{i=0}^m(\frac{10}{i})(\frac{20}{m-i})={\sum }_{i=0}^m{}^{10}{C}_i\times {}^{20}{C}_{m-i}$

$={}^{10}{C}_0\times {}^{20}{C}_m+{}^{10}{C}_1\times {}^{20}{C}_{m-1}+{}^{10}{C}_2\times {}^{20}{C}_{m-2}+\cdot \cdot \cdot +{}^{10}{C}_m$

$={}^{30}{C}_m$ ....... Using the above concept

$\therefore {\sum }_{i=0}^m(\frac{10}{i})(\frac{20}{m-i})={}^{30}{C}_m$

We know that, ${}^n{C}_r$ is maximum when $r=\frac{n}{2}$

$\therefore {}^{30}{C}_m$ is maximum when $m=\frac{30}{2}=15$

Hence, option C is correct.