Find the number of rational terms in the expansion of `(2^((1)/(3)) + 3^((1)/(5)))^(600)`

Answer

To find the number of rational terms in the expansion of \((2^{\frac{1}{3}} + 3^{\frac{1}{5}})^{600}\), we can follow these steps:

### Step 1: Understand the General Term
The general term in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = 2^{\frac{1}{3}}\), \(b = 3^{\frac{1}{5}}\), and \(n = 600\). Therefore, the general term becomes:
\[ T_r = \binom{600}{r} (2^{\frac{1}{3}})^{600-r} (3^{\frac{1}{5}})^{r} \]
This simplifies to:
\[ T_r = \binom{600}{r} 2^{\frac{600 - r}{3}} 3^{\frac{r}{5}} \]

### Step 2: Conditions for Rational Terms
For \(T_r\) to be a rational term, both exponents \(\frac{600 - r}{3}\) and \(\frac{r}{5}\) must be integers. This leads us to the following conditions:

1. \(\frac{600 - r}{3}\) is an integer, which implies \(600 - r \equiv 0 \mod 3\).
2. \(\frac{r}{5}\) is an integer, which implies \(r \equiv 0 \mod 5\).

### Step 3: Solve for \(r\)
From the first condition, we can express \(r\) as:
\[ r \equiv 0 \mod 3 \quad \Rightarrow \quad r = 3k \quad \text{for some integer } k \]
From the second condition:
\[ r \equiv 0 \mod 5 \quad \Rightarrow \quad r = 5m \quad \text{for some integer } m \]

### Step 4: Find Common Values of \(r\)
To satisfy both conditions, \(r\) must be a common multiple of 3 and 5. The least common multiple (LCM) of 3 and 5 is 15. Thus, we can express \(r\) as:
\[ r = 15n \quad \text{for some integer } n \]

### Step 5: Determine the Range of \(r\)
Since \(r\) can vary from 0 to 600, we need to find the values of \(n\) such that:
\[ 0 \leq 15n \leq 600 \]
Dividing through by 15 gives:
\[ 0 \leq n \leq 40 \]
Thus, \(n\) can take values from 0 to 40, inclusive.

### Step 6: Count the Number of Rational Terms
The possible values of \(n\) are:
\[ 0, 1, 2, \ldots, 40 \]
This gives us a total of:
\[ 40 - 0 + 1 = 41 \]

### Conclusion
The number of rational terms in the expansion of \((2^{\frac{1}{3}} + 3^{\frac{1}{5}})^{600}\) is **41**.