The number of irrational terms in the binomial expansion of `(3^(1//5) + 7^(1//3))^(100)` is ___

Answer

To find the number of irrational terms in the binomial expansion of \((3^{1/5} + 7^{1/3})^{100}\), we can follow these steps:

### Step 1: Identify the General Term
The general term (T_r) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = 3^{1/5}\), \(b = 7^{1/3}\), and \(n = 100\). Thus, the general term becomes:
\[ T_r = \binom{100}{r} (3^{1/5})^{100-r} (7^{1/3})^r = \binom{100}{r} 3^{(100-r)/5} 7^{r/3} \]

### Step 2: Determine Conditions for Rational Terms
For \(T_r\) to be rational, both exponents \((100 - r)/5\) and \(r/3\) must be integers. This leads us to the following conditions:
1. \((100 - r)/5\) is an integer \(\Rightarrow 100 - r \equiv 0 \mod 5\)
2. \(r/3\) is an integer \(\Rightarrow r \equiv 0 \mod 3\)

### Step 3: Find Values of \(r\)
From the first condition, \(100 - r\) must be a multiple of 5. Therefore, \(r\) can take values such that:
\[ r = 100, 95, 90, 85, 80, 75, 70, 65, 60, 55, 50, 45, 40, 35, 30, 25, 20, 15, 10, 5, 0 \]
This gives us possible values of \(r\) as \(0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100\).

From the second condition, \(r\) must be a multiple of 3:
\[ r = 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99 \]

### Step 4: Find Common Values of \(r\)
Now we find the common values of \(r\) that satisfy both conditions:
- The multiples of 5 from the first condition are: \(0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100\)
- The multiples of 3 from the second condition are: \(0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99\)

The common multiples of both conditions are:
\[ r = 0, 15, 30, 45, 60, 75, 90 \]
This gives us a total of 7 rational terms.

### Step 5: Calculate Total Terms
The total number of terms in the expansion is given by \(n + 1\):
\[ 100 + 1 = 101 \]

### Step 6: Calculate Irrational Terms
The number of irrational terms is given by:
\[ \text{Number of irrational terms} = \text{Total terms} - \text{Rational terms} = 101 - 7 = 94 \]

### Final Answer
Thus, the number of irrational terms in the binomial expansion of \((3^{1/5} + 7^{1/3})^{100}\) is **94**.

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