Binomial Theorem
Binomial Coefficients
Grade 11
Question:
<p>The value of \({}^nC_0 \times 2^n - {}^nC_1 \times 2^{n-2} + {}^nC_2 \times 2^{n-4}C_r - \ldots\) is equal to</p>
<p>(1) \({}^nC_{r-n} \times 2^{2n-r}\) if \(r \geq n\)</p>
<p>(2) 0, if \(r < n\)</p>
<p>(3) \({}^nC_{r-n} \times 2^{n-r}\) if \(r \geq n\)</p>
<p>(4) \(-{}^nC_{r-n} \times 2^{2n-r}\) if \(r < n\)</p>
Step-by-Step Solution
Key Concept: Recognize this as the real part of the binomial expansion of (1+i)^n where i=√(-1), since multiplying by powers of 2 and alternating signs suggests complex exponential form. Use (1+i)^n = (√2)^n × e^(inπ/4) to extract the real component.
<p><strong>Step 1:</strong> Rewrite the series with pattern recognition:</p><p>C₀·2^n - C₁·2^(n-2) + C₂·2^(n-4) - ... = 2^n[C₀ - C₁·2^(-2) + C₂·2^(-4) - ...]</p><p><strong>Step 2:</strong> Recognize this as the real part of binomial expansion of (1+i)^n:</p><p>(1+i)^n = Σ C_r · i^r</p><p>Where i^r cycles: 1, i, -1, -i, 1, ...</p><p><strong>Step 3:</strong> Convert 1+i to polar form:</p><p>1+i = √2 · e^(iπ/4)</p><p>(1+i)^n = (√2)^n · e^(inπ/4) = 2^(n/2) · [cos(nπ/4) + i·sin(nπ/4)]</p><p><strong>Step 4:</strong> Multiply by 2^(n/2) to match the original expression:</p><p>2^n · [C₀ - C₁·2^(-2) + C₂·2^(-4) - ...] = Re[(1+i)^n · 2^(n/2)]</p><p>= 2^n · cos(nπ/4)</p><p><strong>Step 5:</strong> The answer depends on n mod 8:</p><p>For n≡0(mod 8): 2^n</p><p>For n≡4(mod 8): -2^n</p><p>∴ Answer: <strong>2^(n/2)·cos(nπ/4)</strong> or <strong>2^n·cos(nπ/4)</strong> depending on options given</p>
Correct Answer: A,B