The number of integral terms in the expansion of `(3^(1//2)+5^(1//8))^(256)` is

Options

A. 33
B. 34
C. 35
D. 32

Answer

To find the number of integral terms in the expansion of \((3^{1/2} + 5^{1/8})^{256}\), we can follow these steps:

### Step 1: Identify the General Term
The general term \(T_r\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
For our expression, \(a = 3^{1/2}\), \(b = 5^{1/8}\), and \(n = 256\). Thus, the general term becomes:
\[ T_r = \binom{256}{r} (3^{1/2})^{256 - r} (5^{1/8})^r \]
This simplifies to:
\[ T_r = \binom{256}{r} 3^{(256 - r)/2} 5^{r/8} \]

### Step 2: Determine Conditions for Integral Terms
For \(T_r\) to be an integer, both exponents \((256 - r)/2\) and \(r/8\) must be integers.

1. **Condition for \(3^{(256 - r)/2}\)**:
\[ \frac{256 - r}{2} \text{ is an integer} \implies 256 - r \text{ must be even} \]
Since \(256\) is even, \(r\) must also be even.

2. **Condition for \(5^{r/8}\)**:
\[ \frac{r}{8} \text{ is an integer} \implies r \text{ must be a multiple of } 8 \]

### Step 3: Find Values of \(r\)
Let \(r = 8k\) where \(k\) is an integer. Since \(r\) must be even and a multiple of \(8\), we can express \(r\) as:
\[ r = 0, 8, 16, \ldots, 256 \]
To find the maximum value of \(k\):
\[ 8k \leq 256 \implies k \leq 32 \]
Thus, \(k\) can take values from \(0\) to \(32\).

### Step 4: Count the Integral Terms
The possible values of \(k\) are \(0, 1, 2, \ldots, 32\). This gives us a total of:
\[ 32 - 0 + 1 = 33 \text{ integral terms} \]

### Final Answer
The number of integral terms in the expansion of \((3^{1/2} + 5^{1/8})^{256}\) is \(\boxed{33}\).