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If f(n) = sum(i=0)^(n)(("30"),("30-i"))...

If ` f(n) = sum_(i=0)^(n)(("30"),("30-i"))(("20"),("30-i"))`, then

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The correct Answer is:
a,b,d
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)`
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