Find the number of rational terms in the expansion of
`(9^(1//4)+8^(1//6))^(1000)`.

Answer

To find the number of rational terms in the expansion of \((9^{1/4} + 8^{1/6})^{1000}\), we will follow these steps:

### Step 1: Rewrite the terms in the binomial expression
We can express \(9^{1/4}\) and \(8^{1/6}\) in terms of their prime factors:
\[ 9^{1/4} = (3^2)^{1/4} = 3^{1/2} \]
\[ 8^{1/6} = (2^3)^{1/6} = 2^{1/2} \]
Thus, we can rewrite the expression as:
\[ (3^{1/2} + 2^{1/2})^{1000} \]

### Step 2: Identify the general term in the binomial expansion
The general term \(T_r\) in the expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
For our expression, \(a = 3^{1/2}\), \(b = 2^{1/2}\), and \(n = 1000\). Therefore, the general term becomes:
\[ T_r = \binom{1000}{r} (3^{1/2})^{1000 - r} (2^{1/2})^r = \binom{1000}{r} 3^{(1000 - r)/2} 2^{r/2} \]

### Step 3: Determine when the terms are rational
For \(T_r\) to be a rational term, both exponents \((1000 - r)/2\) and \(r/2\) must be integers. This means that both \(1000 - r\) and \(r\) must be even numbers.

### Step 4: Set conditions for \(r\)
Let \(r = 2k\) where \(k\) is an integer. Then:
\[ 1000 - r = 1000 - 2k \]
For \(1000 - r\) to be even, \(1000\) is already even, thus \(2k\) must also be even, which it is.

### Step 5: Find the range of \(k\)
Now we need to determine the possible values of \(k\):
- Since \(r\) must be between \(0\) and \(1000\), we have:
\[ 0 \leq 2k \leq 1000 \implies 0 \leq k \leq 500 \]
Thus, \(k\) can take values from \(0\) to \(500\).

### Step 6: Count the number of rational terms
The number of integer values \(k\) can take is:
\[ k = 0, 1, 2, \ldots, 500 \]
This gives us a total of \(501\) values.

### Conclusion
The number of rational terms in the expansion of \((9^{1/4} + 8^{1/6})^{1000}\) is \(501\).

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