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If (1+x)^(n) = C(0) + C(1)x + C(2)x^(2) ...

If `(1+x)^(n) = C_(0) + C_(1)x + C_(2)x^(2) + "….." + C_(n)x^(n)`, then `C_(0) - (C_(0) + C_(1)) +(C_(0) + C_(1) + C_(2)) - (C_(0) + C_(1) + C_(2) + C_(3))+ "….."` `(-1)^(n-1) (C_(0) + C_(1) + "……" + C_(n-1))` is (where n is even integer and `C_(r) = .^(n)C_(r)`)

A

a positive value

B

a negative value

C

divisivle by ` 2^(n-1)`

D

divisible by ` 2^(n)`

Text Solution

Verified by Experts

The correct Answer is:
b,c
We have , `C_(0) - (C_(0) + C_(1)) + (C_(0) + C_(1) + C_(2)) - (C_(0) `
` + C_(1) + C_(2) + C_(3) ) + … (-1)^(n-1) (C_(0) + C_(1) + …+ C_(n-1))`
For even integer , take n = 2 m, we get
` = C_(0)- (C_(0) + C_(1)) + (C_(0) + C_(1) + C_(2))`
` - (C_(0) + C_(1) + C_(2) + C_(3)) + ...- (C_(0) + C_(1) + ....+ C_(2m-1))`
` = - (C_(1) + C_(3) + C_(5) + ...+ C_(2m-1))`
` = - (C_(1) + C_(3) + C_(5) + ...+ C_(n-1))" " [ because n = 2 m]`
`= - 2^(n-1)`
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