<p>The number of integer terms in the expansion of \((\sqrt{3} + \sqrt[8]{5})^{256}\) is:</p>
Step-by-Step Solution
Key Concept: An integer term occurs when both exponents of √3 and ∜[8]{5} yield integers; this requires the exponent of 3^(1/2) to be even and the exponent of 5^(1/8) to be divisible by 8.
<p><strong>Step 1:</strong> Write the general term using binomial theorem:</p><p>T_{r+1} = C(256,r)·(√3)^(256-r)·(∜[8]{5})^r = C(256,r)·3^((256-r)/2)·5^(r/8)</p><p><strong>Step 2:</strong> For T_{r+1} to be an integer, both exponents must be integers:</p><p>• (256-r)/2 must be an integer ⟹ (256-r) must be even ⟹ r must be even</p><p>• r/8 must be an integer ⟹ r must be divisible by 8</p><p><strong>Step 3:</strong> Find values of r satisfying both conditions:</p><p>r must be divisible by 8 (which automatically makes it even) with 0 ≤ r ≤ 256</p><p>Valid values: r ∈ {0, 8, 16, 24, ..., 256}</p><p><strong>Step 4:</strong> Count the terms:</p><p>These are r = 8k where k = 0, 1, 2, ..., 32</p><p>Number of terms = 32 + 1 = <strong>33</strong></p><p>∴ Answer: B</p>
Correct Answer: B