If `f(m) = sum_(i=0)^m ((30),(30-i)) ((20),(m-i))` where `((p),(q)) = pCq,` then (A) maximum value of f(m) is 50C 25 (B) `f(0)+f(1) +...+f(50) =2^50` (C) f(m) is always divisible by 50 (`1<=m<=49`) (D)`sum_(m=0)^(50) (f(m))^2=100C50`

Options

A. maximum value of (n) is ` ""^(50)C_(25)`
B. ` f(0) + f(1) + f(2) + …+ f(50) = 2^(50)`
C. f (n) is always divisible by 50
D. ` f^(2) (0) + f^(2)(1) + f^(2)(2) + …+ f^(2) (50) = ""^(100)C_(50)`

Answer

a,b,d

Explanation

We have ,
`f(n) = sum_(i=0)^(n) ((30),(30-i))((20),(n-i)) = sum_(i=0)^(n) ((30)/(i)) ((20)/(n-i)) = ""^(50)C_(n)`
` therefore ` f(n) is greatest , when n = 25
` therefore ` Maximum value of f(n) is ` ""^(50)C_(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_(n)` is not divisible by 50 for any n as 50 is not a prime number .
` sum_(i=0)^(n) (f(n))^(2) = (""^(50)C_(0))^(2) + (""^(50)C_(1))^(2) + ...+ (""^(50)C_(50))^(2) = ""^(100)C_(50)`