The number of integer terms in the expansion of `(5^(1//2)+7^(1//6))^(642)` is

Options

A. 105
B. 108
C. 106
D. 109

Answer

To find the number of integer terms in the expansion of \((5^{1/2} + 7^{1/6})^{642}\), we will follow these steps:

### Step 1: Identify the General Term
Using the Binomial Theorem, the general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, \(a = 5^{1/2}\), \(b = 7^{1/6}\), and \(n = 642\). Therefore, the general term becomes:
\[ T_{r+1} = \binom{642}{r} (5^{1/2})^{642 - r} (7^{1/6})^r \]
This simplifies to:
\[ T_{r+1} = \binom{642}{r} 5^{(642 - r)/2} 7^{r/6} \]

### Step 2: Determine Conditions for Integer Terms
For \(T_{r+1}\) to be an integer, both \(5^{(642 - r)/2}\) and \(7^{r/6}\) must be integers. This implies:
1. \((642 - r)/2\) must be a non-negative integer, meaning \(642 - r\) must be even.
2. \(r/6\) must also be a non-negative integer, meaning \(r\) must be a multiple of 6.

### Step 3: Analyze the Conditions
From the first condition, since \(642\) is even, \(r\) must also be even for \(642 - r\) to be even. The second condition states that \(r\) must be a multiple of 6. Thus, we can express \(r\) as:
\[ r = 6k \quad \text{for some integer } k \]
Substituting this into the condition that \(r\) must be less than or equal to \(642\):
\[ 6k \leq 642 \implies k \leq \frac{642}{6} = 107 \]
Thus, \(k\) can take values from \(0\) to \(107\).

### Step 4: Count the Possible Values of \(k\)
The possible values of \(k\) are:
\[ k = 0, 1, 2, \ldots, 107 \]
This gives us a total of:
\[ 107 - 0 + 1 = 108 \text{ values} \]

### Conclusion
Therefore, the number of integer terms in the expansion of \((5^{1/2} + 7^{1/6})^{642}\) is:
\[ \boxed{108} \]