Question

The sum $\sum _{i=0}^m(\frac{10}{i})(\frac{20}{m-i})$ be maximum when m is

• A. 15
• B. 5
• C. 10
• D. 20

Answer 1

The correct option is A 15

Add upper suffixces if their sum is even the half of it will be the value r otherwise the value of r be sum of upper suffixces +1 divided by 2.
Given $S=\sum _{t=0}^m(\frac{10}{i})(\frac{20}{m-i})=(\frac{10}{0})(\frac{20}{m})+(\frac{10}{1})(\frac{20}{m-1})+(\frac{10}{2})(\frac{20}{m-2})+...+(\frac{10}{m})(\frac{20}{0})$
In such cases we observe the upper suffixces and lower suffixces.
Upper suffices are 10 and 20 and lower suffixces taken the value (0...m).
In such cases we consider the two Binomial expansions ${(1+x)}^p$ and ${(1+x)}^q$ (Here p=10, q=20 or q=10, p=20) multiply them and collect the required exponent of x to which the given expression can be obtained.
Consider ${(1+x)}^{10}{(1+x)}^{20}=({}^{10}{C}_0{+}^{10}{C}_2{x}^2+...{+}^{10}{C}_{{10}^{x^{10}}})({}^{20}{C}_0{+}^{20}{C}_{1}x+...{+}^{20}{C}_{{20}^{x^{20}}})$
${(1+x)}^{30}=({}^{10}{C}_0{+}^{10}{C}_{1^{x^1}}+...{+}^{10}{C}_{{10}^{x^{10}}}).({}^{20}{C}_0{+}^{20}{C}_1{x}^1+...{+}^{20}{C}_{m^{x^m}}+...{+}^{20}{C}_{{20}^{x^{20}}})$
and collecting coefficient of $x^m$ both side, we get ${}^{30}{C}_m=(\frac{10}{0})(\frac{20}{m})+(\frac{10}{1})(\frac{20}{m-2})+...+(\frac{10}{m})(\frac{20}{0})$ now for max value of the expression will be at $m=\frac{30}{2}=15$ as upper suffix is even $\therefore m=15$ which is given in (a).