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If (1 + x)^n = C0+C1x+C2x^2 + ... + Cnx^...

If `(1 + x)^n = C_0+C_1x+C_2x^2 + ... + C_nx^n`,then for n odd,`C_0^2-C_1^2+C_2^2-C_3^2 + ... +(-1) C_n^2`, is equal to

A

`.^(2n)C_(n)`

B

`(-1)^(n//2).^(n)C_(n//2)`

C

0

D

`.^(2n)C_(n-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+…..+C_(1)x^(n)+….+C_(n)X^(n)`
& `(x-1)^(n)=C_(0)x^(n)-C_(1)x^(n-1)+C_(2)x^(n-2)+……+(-1)^(n)C_(n)`
Multiplying and compare coefficient of `x^(n)`
`C_(o)^(2)-C_(1)^(2)+C_(2)^(2)-C_(3)^(2)+…..+(-1)^(n)C_(n)^(2)=(-1)^(n//2) .^(n)C_(n//2)`
If n is odd then
`C_(o)^(2)-C_(1)^(2)+C_(2)^(2)-C_(3)^(2)+....+(-1)^(n)C_(n)^(2)=0`
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