If `f(m) = sum_(i=0)^m ({:(30),(30-i):})({:(20),(m-i):})` where `({:(p),(q):})= ""^(p)C_(q)`, then

Options

A. A. maximum value of `f(m)` is `.^(50)C_(25)`
B. B. `f(0) + f(1)+"….."+f(50) = 2^(50)`
C. C. `f(m)` is always divisible by `50(1 le m le 49)`
D. D. The value of `underset(m=0)overset(50)sum(f(m))^(2) = .^(100)C_(50)`

Answer

A, B, D

Explanation

`f(m)= underset(i=0)overset(m)sum({:(" "30 ),(30-i):})({:(" "30 ),(m-i):})= underset(i=0)overset(m)sum({:(" "30 ),(" "i):})({:(" "20 ),(m-i):})= .^(50)C_(m)`
`f(m)` is greatest when `m = 25`. Also,
`f(0)+f(1)+"….." +f(50)= .^(50)C_(0) + .^(50)C_(1)+.^(50)C_(2)+"....."+.^(50)C_(50) = 2^(50)`
Also, `.^(50)C_(m)`is not divisible by 50 for all m as 50 is not a prime number
`underset(m=0)overset(50)sum(f(m))^(2)=(.^(50)C_(0))^(2)+(.^(50)C_(1))^(2)+(.^(50)C_(2))^(2)+"......."+(.^(50)C_(50))^(2)`
`= .^(100)C_(50)`