Question

The sum ${\sum }_{i=0}^m(\frac{10}{i})(\frac{20}{m-i}),where(\frac{p}{q})=0ifp>q,$ is maximum when m is equal to

• A. 5
• B. 10
• C. 15
• D. 20

Answer 1

The correct option is C

15

${\sum }_{i=0}^m(\frac{10}{i})(\frac{20}{m-i})$ is the coefficient of $x^m$ in the expansion of $(1+x{)}^{10}(x+1{)}^{20}$,
$\Rightarrow {\sum }_{i=0}^m(\frac{10}{i})(\frac{20}{m-i})$ is the coefficient of $x^m$ in the expansion of $(1+x{)}^{30}$
$i.e.{\sum }_{i=0}^m(\frac{10}{i})(\frac{20}{m-i}){=}^{30}{C}_m=(\frac{30}{m})$ . . . (i)
and we know that, $(\frac{n}{r})$ is maximum, when
${(\frac{n}{r})}_{max}=\{\begin{array}{cc}r=\frac{n}{2}& ifnϵeven\\ r=\frac{n\pm 1}{2}& ifnϵodd\end{array}$
Hence, $(\frac{30}{m})$ is maximum when m = 15