<p><strong>35.</strong> If \((1+x)(1+x^2)(1+x^4)\cdots(1+x^{128}) = \displaystyle\sum_{r=0}^{n} x^r\), then \(n\) is equal to</p>
Step-by-Step Solution
Key Concept: Multiply the product by (1-x) to telescope the left side into (1-x^256)/(1-x), which reveals the sum contains all powers from x^0 to x^255. The highest power determines n.
<p><strong>Step 1:</strong> Recognize the pattern in exponents: 1, 2, 4, 8, ..., 128 are powers of 2, i.e., 2^0, 2^1, 2^2, ..., 2^7.</p><p><strong>Step 2:</strong> Multiply both sides by (1-x):</p><p>(1-x)(1+x)(1+x²)(1+x⁴)···(1+x^128) = (1-x)∑_{r=0}^{n} x^r</p><p><strong>Step 3:</strong> Apply difference of squares repeatedly on the left side:</p><p>(1-x)(1+x) = (1-x²)</p><p>(1-x²)(1+x²) = (1-x⁴)</p><p>(1-x⁴)(1+x⁴) = (1-x⁸)</p><p>Continuing this pattern: (1-x^256)</p><p><strong>Step 4:</strong> The left side telescopes to (1-x^256), so:</p><p>(1-x^256) = (1-x)∑_{r=0}^{n} x^r</p><p><strong>Step 5:</strong> Expand the right side using the geometric series formula:</p><p>(1-x)∑_{r=0}^{n} x^r = (1+x+x²+···+x^n) - (x+x²+···+x^{n+1}) = 1-x^{n+1}</p><p><strong>Step 6:</strong> Equate: 1-x^256 = 1-x^{n+1}</p><p>Therefore: n+1 = 256, so <strong>n = 255</strong></p><p>∴ Answer: B</p>
Correct Answer: B