InterviewSolution
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InterviewSolution
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If for `z`as real or complex, `(1+z^2+z^4)^8=C_0+C1z2+C2z4++C_(16)z^(32)t h e n``C_0-C_1+C_2-C_3++C_(16)=1``C_0+C_3+C_6+C_9+C_(12)+C_(15)=3^7``C_2+C_5+C_6+C_(11)+C_(14)=3^6``C_1+C_4+C_7+C_(10)+C_(13)+C_(16)=3^7`A. `C_(0) - C_(1) + C_(2) - C_(3) + "….." + C_(16) = 1`B. `C_(0) + C_(3) + C_(6) + C_(12) + C_(15) = 3^(7)`C. `C_(2) + C_(5) + C_(8) + C_(11) + C_(14) = 3^(6)`D. `C_(1) + C_(4) + C_(7) + C_(10) + C_(13) + C_(16) = 3^(7)` |
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Answer» Correct Answer - A::B::D `(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)` |
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