The number of terms in the expansion of `(sqrt(5)+11^(1//4))^(124)` which are integers is

Options

A. 30
B. 31
C. 32
D. 0

Answer

To find the number of integer terms in the expansion of \((\sqrt{5} + 11^{1/4})^{124}\), we can use the Binomial Theorem. The general term in the expansion can be expressed as follows:

### Step 1: Write the General Term
The general term \(T_{r+1}\) in the expansion of \((x + a)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} a^r \]
In our case, \(x = \sqrt{5}\), \(a = 11^{1/4}\), and \(n = 124\). Thus, the general term becomes:
\[ T_{r+1} = \binom{124}{r} (\sqrt{5})^{124-r} (11^{1/4})^r \]
This simplifies to:
\[ T_{r+1} = \binom{124}{r} (5^{1/2})^{124-r} (11^{r/4}) = \binom{124}{r} 5^{(124-r)/2} 11^{r/4} \]

### Step 2: Conditions for Integer Terms
For \(T_{r+1}\) to be an integer, both \(5^{(124-r)/2}\) and \(11^{r/4}\) must be integers.

1. **Condition for \(5^{(124-r)/2}\)**:
\((124 - r)/2\) must be an integer, which implies that \(124 - r\) must be even. Therefore, \(r\) must be even.

2. **Condition for \(11^{r/4}\)**:
\(r/4\) must also be an integer, which implies that \(r\) must be a multiple of 4.

### Step 3: Finding Valid Values for \(r\)
Since \(r\) must be both even and a multiple of 4, we can express \(r\) as:
\[ r = 4k \quad \text{for some integer } k \]
Now, we need to find the range of \(k\) such that \(0 \leq r \leq 124\):
\[ 0 \leq 4k \leq 124 \implies 0 \leq k \leq 31 \]
Thus, \(k\) can take values from 0 to 31, which gives us 32 possible values for \(k\).

### Step 4: Conclusion
Therefore, the number of integer terms in the expansion of \((\sqrt{5} + 11^{1/4})^{124}\) is:
\[ \boxed{32} \]