If for z as real or complex- 1- z 2 - z 4 8 - C 0 - C 1 z 2 - C 2 z 2 - C 16 z 32- then-

The correct option is C C _ 0 C _ 1 + C _ 2 C _ 3 +··· + C _ 16 =1 Let z=i Then z^2= 1 z^4=1 Therefore z^2(2n 1)= 1 and z^4n= ...

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Fundamental Laws of Logarithms

If for z as...

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If for $z$ as real or complex,
${(1 + z^{2} + z^{4})}^{8} = C_{0} + C_{1} z^{2} + C_{2} z^{2} \cdot \cdot \cdot + C_{16} z^{32}$ , then:

C_{0} - C_{1} + C_{2} - C_{3} + \cdot \cdot \cdot + C_{16} = 1

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C_{0} - C_{1} + C_{2} - C_{3} + \cdot \cdot \cdot + C_{16} = 0

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C_{0} + C_{1} + C_{2} + C_{3} + \cdot \cdot \cdot + C_{16} = 1

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None of these

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z

(1 + z^{2} + z^{4})^{8} = C_{0} + C_{1} z^{2} + C_{2} z^{4} + . . . + C_{16} z^{32}

z

{(1 + z^{2} + z^{4})}^{8} = C_{0} + C_{1} z^{2} + C_{2} z^{4} \cdot \cdot \cdot \cdot \cdot \cdot + C_{16} z^{32}

C_{0} + C_{3} + . . . . + C_{15} + (C_{2} + C_{5} + C_{8} + C_{11} + C_{14}) ω +

(C_{1} + C_{4} + C_{7} + . . . . + C_{16}) ω^{2}

ω

C_{0} + C_{1} + C_{2} + \dots \dots . + C_{n} = 128

C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \frac{C_{3}}{4} + \dots \dots . =

\frac{C_{1}}{C_{0}} + 2. \frac{C_{2}}{C_{1}} + 3. \frac{C_{3}}{C_{2}} + . . . . . + n . \frac{C_{n}}{C_{n - 1}}

{(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}

C_{0} C_{2} + C_{1} C_{3} + C_{2} C_{4} + . . . + C_{n - 2} C_{n} =

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