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

If ` (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + C_(3)x^(3) + …+ 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_(2) + ...+ C_(n-1)) ` , when n is even integer is

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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