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The sum S = ""^(20)C(2) + 2*""^(20)C(3) ...

The sum S = `""^(20)C_(2) + 2*""^(20)C_(3) + 3 *""^(20)C_(4) + ...+ 19 * ""^(20)C_(20) ` is equal to

A

` 1 + 5 *2 ^(20)`

B

`1 + 2^(21)`

C

`1 + 9 * 2^(20)`

D

`2^(20)`

Text Solution

Verified by Experts

The correct Answer is:
c
We have , ` (1 + x)^(20) = ""^(20)C_(0) + ""^(20)C_(1) x + ""^(20)C_(2) x^(2) x^(2) + …+ ""^(20)C_(20) x^(20)`
On dividing by x , we get
`((1+ x)^(20))/(x)= ( ""^(20)C_(0))/(x) + ""^(20)C_(1) + ""^(20)C_(2)x + ""^(20)C_(3) x^(2) + ….+ ""^(20)C_(20) x^(19)`
On differentiting w.r.t.x, we get
`(20(1 + x)^(19) *x- (1+ x)^(20))/(x^(2)) = (""^(20)C_(0))/(x^(2)) + 0 + ""^(20)C_(2)+ 2*""^(20)C_(3)x + ...+ 19 * ""^(20)C_(20) x^(18)`
On putting x = 1 , we get
`20(2)^(19) - (2)^(20) = - (1)/(1) + ""^(20)C_(2) + 2*""^(20)C_(3)x + ...+ 19 * ""^(20)C_(20) `
`therefore ""^20C_(2) + 2*""^(20)C_(3) + ...+ 19* ""^(20)C_(20) = 1 + 10*2^(20) - 2^(20) = 1+ 9*2^(20)`.
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