Number of integral term in the expansion of `(7^(1/2)+11^(1/6))^(824)`

Options

A. 139
B. 138
C. 140
D. 137

Answer

To find the number of integral terms in the expansion of \((7^{1/2} + 11^{1/6})^{824}\), we will follow these steps:

### Step 1: Understand the Binomial Expansion
The binomial expansion of \((a + b)^n\) can be expressed as:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
where \(T_r\) is the \(r^{th}\) term, \(n\) is the exponent, \(a\) and \(b\) are the terms being expanded.

### Step 2: Identify \(a\), \(b\), and \(n\)
In our case:
- \(a = 7^{1/2}\)
- \(b = 11^{1/6}\)
- \(n = 824\)

Thus, the \(r^{th}\) term in the expansion will be:
\[ T_r = \binom{824}{r} (7^{1/2})^{824 - r} (11^{1/6})^r = \binom{824}{r} 7^{(824 - r)/2} 11^{r/6} \]

### Step 3: Conditions for Integral Terms
For \(T_r\) to be an integral term, both exponents \((824 - r)/2\) and \(r/6\) must be integers.

1. **Condition from \(7^{(824 - r)/2}\)**:
- \((824 - r)/2\) is an integer \(\Rightarrow 824 - r\) must be even.
- Since \(824\) is even, \(r\) must also be even.

2. **Condition from \(11^{r/6}\)**:
- \(r/6\) is an integer \(\Rightarrow r\) must be a multiple of \(6\).

### Step 4: Find Possible Values of \(r\)
Since \(r\) must be both even and a multiple of \(6\), we can express \(r\) as:
\[ r = 6k \quad \text{for integers } k \]

### Step 5: Determine the Range of \(k\)
We need \(r\) to be less than or equal to \(824\):
\[ 6k \leq 824 \Rightarrow k \leq \frac{824}{6} \approx 137.33 \]
Thus, \(k\) can take integer values from \(0\) to \(137\).

### Step 6: Count the Values of \(k\)
The possible values of \(k\) are:
\[ k = 0, 1, 2, \ldots, 137 \]
This gives us \(138\) possible values for \(k\).

### Step 7: Conclusion
Therefore, the total number of integral terms in the expansion of \((7^{1/2} + 11^{1/6})^{824}\) is:
\[ \boxed{138} \]