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[(^nC0+^n C3+)-1/2(^n C1+^n C2+^n C4+^n ...

`[(^nC_0+^n C_3+)-1/2(^n C_1+^n C_2+^n C_4+^n C_5]^2 + 3/4(^n C_1-^n C_2+^n C_4 +)^2=` a.`3` b. `4` c. `2` d. `1`

A

3

B

4

C

2

D

1

Text Solution

Verified by Experts

The correct Answer is:
D
`(1+omega)^(n) = .^(n)C_(0) + .^(n)C_(1) +omega+"....."`
`= (.^(n)C_(0)+.^(n)C_(3) +"....")+(.^(n)C_(1)+.^(n)C_(4)+".....")((-1+sqrt(3)i)/(2))+(.^(n)C_(2)+.^(n)C_(5)+"....")((-1-sqrt(3)t)/(2))`
`= (.^(n)C_(0) + .^(n)C_(3) + "....") -1/2 (.^(n)C_(1) + .^(n)C_(2) + .^(n)C_(4) + .^(n)C_(5)".....")+(isqrt(3))/(2)(.^(n)C_(1)-.^(n)C_(2)+.^(n)C_(4)-.^(n)C_(5) + "....")`
Equating the modulus, we get `|(-omega^(2))^(n)| = 1`.
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CENGAGE ENGLISH-BINOMIAL THEOREM-Single Correct Answer
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