If for `z` as real or complex, `(1+z^2+z^4)^8=C_0+C_1z^2+C_2z^4+...+C_(16)z^(32) then ` prove that `C_(0) - C_(1) + C_(2) - C_(3) + "….." + C_(16) = 1` and `C_(0) + C_(3) + C_(6) + C_(12) + C_(15) = 3^(7)`

Answer

A, B, D

Explanation

`(1+z^(2)+z^(4))^(8) = C_(0) + C_(1)z^(2) + X_(2)z^(4) + "….." + C_(16)z^(32) " "(1)`
Putting `x = i`, where `i = sqrt(-1)`.
`(1-1+1)^(8) = C_(0) - C_(1) + C_(2) - C_(3) + "….." + C_(16)`
or `C_(0) - C_(1) + C_(2) - C_(3) + "……" + C_(16) = 1`
Also, putting `z = omega`
`(1+omega^(2)+omega^(4))^(8)= C_(0) + C_(1)omega^(2) + C_(2)omega^(4) + "....." + C_(16)omega^(32)`
or `C_(0) + C_(1)omega^(2) + C_(2)omega + C_(3) + "...." + C_(16)omega^(2) = 0 " "(3)`
Putting `x = omega^(2)`.
`(1+omega^(4)+omega^(8))^(8) = C_(0) + C_(1)omega^(4) + C_(2)omega^(8) + "......" + C_(16)omega^(64)`
or `C_(0) + C_(1)omega + C_(2)omega^(2) + "....." + C_(16)omega = 0 " "(3)`
Putting `x = 1`,
`3^(8) = C_(0) + C_(1) + C_(2) + "....."+C_(16) " "(4)`
Adding (2), (3) and (4), we have
`3(C_(0) + C_(3) + "......" + C_(15)) = 3^(8)`
or `C_(0) + C_(3) + "......" + C_(15) = 3^(7)`
Similarly, first multiplying (1) by x and then putting `1 omega, omega^(2)` and adding, we get
`C_(1) + C_(4) + C_(7) + C_(10) + C_(13)+ C_(16) = 3^(7)`
Multiplying (1) by `z^(2)` and then putting `1, omega, omega^(2)` and adding, we get
`C_(2)+C_(5)+C_(8)+C_(11)+C_(14)= 3^(7)`