The number of irrational terms in the expansion of `(2^(1//5) + 3^(1//10))^(55)` is

Options

A. 47
B. 56
C. 50
D. 48

Explanation

To find the number of irrational terms in the expansion of \((2^{1/5} + 3^{1/10})^{55}\), we can follow these steps:

### Step 1: Determine the total number of terms in the expansion
Using the binomial theorem, the number of terms in the expansion of \((A + B)^n\) is given by \(n + 1\). Here, \(n = 55\).

\[ \text{Total number of terms} = 55 + 1 = 56 \]

**Hint:** Remember that the number of terms in a binomial expansion is always one more than the exponent.

### Step 2: Identify the general term in the expansion
The general term \(T_r\) in the expansion of \((2^{1/5} + 3^{1/10})^{55}\) can be expressed as:

\[ T_r = \binom{55}{r} (2^{1/5})^{55-r} (3^{1/10})^r = \binom{55}{r} 2^{(55-r)/5} 3^{r/10} \]

**Hint:** The general term combines the contributions from both parts of the binomial raised to the appropriate powers.

### Step 3: Determine when the terms are irrational
For \(T_r\) to be irrational, both \(2^{(55-r)/5}\) and \(3^{r/10}\) must be irrational.

1. **For \(2^{(55-r)/5}\)** to be irrational:
- \((55 - r)/5\) must not be an integer, which occurs when \(55 - r\) is not a multiple of 5. Thus, \(r\) must be such that \(r \equiv 0 \mod 5\).

2. **For \(3^{r/10}\)** to be irrational:
- \(r/10\) must not be an integer, which occurs when \(r\) is not a multiple of 10.

**Hint:** Check the conditions for irrationality carefully by considering the divisibility of \(r\).

### Step 4: Find valid values of \(r\)
Now we need to find values of \(r\) that satisfy both conditions:
- \(r \equiv 0 \mod 5\) (i.e., \(r\) can be 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55)
- \(r \equiv 0 \mod 10\) (i.e., \(r\) can be 0, 10, 20, 30, 40, 50)

The common values of \(r\) that satisfy both conditions are:
- \(r = 0, 10, 20, 30, 40, 50\)

**Hint:** List out the multiples of 5 and 10 to find their intersection.

### Step 5: Count the number of irrational terms
The valid values of \(r\) are \(0, 10, 20, 30, 40, 50\), which gives us a total of 6 values.

### Step 6: Calculate the number of irrational terms
The total number of irrational terms is given by:

\[ \text{Number of irrational terms} = \text{Total terms} - \text{Rational terms} \]

Since we found 6 values of \(r\) that lead to irrational terms, the number of rational terms is \(56 - 6 = 50\).

Thus, the number of irrational terms is:

\[ \text{Number of irrational terms} = 56 - 6 = 50 \]

### Final Answer
The number of irrational terms in the expansion of \((2^{1/5} + 3^{1/10})^{55}\) is **50**.

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