Question

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

• A. $5$
• B. $10$
• C. $15$
• D. $20$

Answer 1

The correct option is C $15$

$\sum _{i=0}^m\left(\begin{array}{c}10\\ i\end{array}\right)\left(\begin{array}{c}20\\ m-i\end{array}\right)$ is the coefficient of $x^m$ in the expansion of ${(1+x)}^{10}{(x+1)}^{20}.$

$\sum _{i=0}^m\left(\begin{array}{c}10\\ i\end{array}\right)\left(\begin{array}{c}20\\ m-i\end{array}\right)$ is the coefficient of $x^m$ in the expansion of ${(1+x)}^{30}$

i.e., $\sum _{i=0}^m\left(\begin{array}{c}10\\ i\end{array}\right)\left(\begin{array}{c}20\\ m-i\end{array}\right)={.}^{30}{C}_m=\left(\begin{array}{c}30\\ m\end{array}\right)$ ...(i)

and we know $\left(\begin{array}{c}n\\ r\end{array}\right)$ is maximum, when ${\left(\begin{array}{c}n\\ r\end{array}\right)}_{max}=\{\begin{array}{c}\begin{array}{cc}r=\frac{n}{2},& n\in \end{array}\\ \begin{array}{cc}r=\frac{n\pm 1}{2},& n\in \end{array}\end{array}$

$\therefore \left(\begin{array}{c}30\\ m\end{array}\right)$ is maximum when $m=15.$