<p>The number of integral terms in the expansion of \(\left(\sqrt{3} + \sqrt[8]{5}\right)^{256}\) is</p>
Step-by-Step Solution
Key Concept: A term in the binomial expansion is integral only when both exponents of the irrational bases become integers simultaneously. Use the general term formula and find which values of r make both 3^((256-r)/2) and 5^(r/8) have integer exponents.
<p><strong>Step 1:</strong> Write the general term in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$:</p><p>$$T_{r+1} = \binom{256}{r}(\sqrt{3})^{256-r}(\sqrt[8]{5})^r = \binom{256}{r} \cdot 3^{\frac{256-r}{2}} \cdot 5^{\frac{r}{8}}$$</p><p><strong>Step 2:</strong> For $T_{r+1}$ to be integral, both exponents must be non-negative integers:</p><p>• For $3^{\frac{256-r}{2}}$: We need $\frac{256-r}{2} \in \mathbb{Z}^+ \cup \{0\}$ → $(256-r)$ is even → $r$ is even</p><p>• For $5^{\frac{r}{8}}$: We need $\frac{r}{8} \in \mathbb{Z}^+ \cup \{0\}$ → $r$ is divisible by 8</p><p><strong>Step 3:</strong> Find values of $r$ satisfying both conditions:</p><p>$r$ must be divisible by 8 (this automatically makes it even)</p><p>So $r \in \{0, 8, 16, 24, ..., 256\}$</p><p><strong>Step 4:</strong> Count the values:</p><p>$r = 8k$ where $k = 0, 1, 2, ..., 32$</p><p>Number of integral terms = 33</p><p>∴ Answer: B</p>
Correct Answer: B