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 will follow these steps:

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the expansion of \( (x + y)^n \) is given by:

\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]

In our case, \( n = 600 \), \( x = 2^{\frac{1}{3}} \), and \( y = 3^{\frac{1}{5}} \). Therefore, the general term becomes:

\[ T_{r+1} = \binom{600}{r} (2^{\frac{1}{3}})^{600-r} (3^{\frac{1}{5}})^r \]

### Step 2: Simplify the General Term
Now we simplify the general term:

\[ T_{r+1} = \binom{600}{r} \cdot 2^{\frac{600-r}{3}} \cdot 3^{\frac{r}{5}} \]

### Step 3: Determine Conditions for Rational Terms
For \( T_{r+1} \) to be a rational number, both exponents \( \frac{600-r}{3} \) and \( \frac{r}{5} \) must be integers. This means:

1. \( 600 - r \) must be divisible by 3.
2. \( r \) must be divisible by 5.

### Step 4: Set Up Equations
From the first condition, we can express \( r \) as:

\[ 600 - r \equiv 0 \mod 3 \implies r \equiv 0 \mod 3 \]

From the second condition, we have:

\[ r \equiv 0 \mod 5 \]

### Step 5: Find Common Multiples
We need \( r \) to be a multiple of both 3 and 5. The least common multiple of 3 and 5 is 15. Therefore, \( r \) must be a multiple of 15.

### Step 6: List Possible Values of \( r \)
The possible values of \( r \) that satisfy \( r = 15k \) for integers \( k \) must also be less than or equal to 600. The multiples of 15 up to 600 are:

\[ 0, 15, 30, 45, \ldots, 600 \]

### Step 7: Count the Terms
To find how many terms there are, we can use the formula for the \( n \)-th term of an arithmetic progression:

\[ a_n = a + (n-1)d \]

Where:
- \( a = 0 \) (the first term)
- \( d = 15 \) (the common difference)
- The last term \( a_n = 600 \)

Setting up the equation:

\[ 600 = 0 + (n-1) \cdot 15 \]

Solving for \( n \):

\[ 600 = (n-1) \cdot 15 \implies n-1 = \frac{600}{15} = 40 \implies n = 41 \]

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

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