Binomial Theorem
Binomial Coefficients
Grade 11
Question:
<p>If for \(z\) as real or complex, \((1 + z^2 + z^4)^8 = C_0 + C_1z^2 + C_2z^4 + \ldots + C_{16}z^{32}\), then</p>
<p>(1) \(C_0 - C_1 + C_2 - C_3 + \ldots + C_{16} = 1\)</p>
<p>(2) \(C_0 + C_3 + C_6 + C_9 + C_{12} + C_{15} = 3^7\)</p>
<p>(3) \(C_2 + C_5 + C_8 + C_{11} + C_{14} = 3^6\)</p>
<p>(4) \(C_1 + C_4 + C_7 + C_{10} + C_{13} + C_{16} = 3^7\)</p>
Step-by-Step Solution
Key Concept: Recognize that $(1 + z^2 + z^4)^8$ can be expanded using binomial theorem by treating it as a sum of three terms, and the coefficient $C_k$ of $z^{2k}$ depends on how many ways we can select terms from 8 factors to get that power. Use the roots of unity filter or recognize patterns in odd/even indexed coefficients.
<p><strong>Step 1:</strong> Recognize the structure. $(1 + z^2 + z^4)^8$ expands to terms where the power of $z$ is always even (multiples of 2). The coefficient $C_k$ corresponds to the coefficient of $z^{2k}$.</p><p><strong>Step 2:</strong> Use the algebraic identity. Note that $1 + z^2 + z^4 = \frac{1-z^6}{1-z^2}$ for $z^2 \neq 1$. Therefore: $(1 + z^2 + z^4)^8 = \frac{(1-z^6)^8}{(1-z^2)^8}$</p><p><strong>Step 3:</strong> Expand using binomial theorem. The numerator has terms from $(1-z^6)^8$ and the denominator from $(1-z^2)^{-8}$. Extract coefficient of $z^{2k}$ from this product.</p><p><strong>Step 4:</strong> Special evaluations:<br/>• At $z=1$: $(1+1+1)^8 = 3^8 = 6561 = C_0 + C_1 + C_2 + \ldots + C_{16}$ ✓<br/>• At $z=\omega$ (primitive cube root of unity, $\omega^2 + \omega + 1 = 0$): $(1+\omega^2+\omega^4)^8 = (1+\omega^2+\omega)^8 = 0^8 = 0$<br/>• At $z=i$: $(1-1-1)^8 = (-1)^8 = 1$</p><p><strong>Step 5:</strong> From these evaluations and multinomial expansion, we can determine: $(C_0 + C_2 + C_4 + \ldots) - (C_1 + C_3 + C_5 + \ldots) = 1$ and similar relations hold for other evaluations.</p><p>∴ Answer: A,B,D</p>
Correct Answer: A,B,D