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If f(m)=sum(i=0)^m(30(^) 30-i)(20(^)m-i)...

If `f(m)=sum_(i=0)^m(30(^) 30-i)(20(^)m-i)w h e r e(p q)=^p C_q ,t h e n` (a) maximum value of `f(m)i s^(50)C_(25)` (b)`f(0)+f(1)+...f(50)=2^(50)` (c)`f(m)` is always divisible by `50(1lt=mlt=49)` (d)The value of `sum_(m=0)^(50)(f(m))^2=^(100)C_(50)`

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)`

Text Solution

Verified by Experts

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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