In the expansion of `(y^(1//5)+x^(1//10))^55`, the number of terms free of radical sign are

Options

A. 5
B. 6
C. 50
D. 56

Answer

To find the number of terms free of radical signs in the expansion of \((y^{1/5} + x^{1/10})^{55}\), we can follow these steps:

### Step 1: Identify the General Term
In the expansion of \((a + b)^n\), the general term is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, \(a = y^{1/5}\), \(b = x^{1/10}\), and \(n = 55\). Therefore, the general term becomes:
\[ T_{r+1} = \binom{55}{r} (y^{1/5})^{55-r} (x^{1/10})^r \]

### Step 2: Simplify the General Term
Now, simplifying the general term:
\[ T_{r+1} = \binom{55}{r} y^{(55-r)/5} x^{r/10} \]
This can be rewritten as:
\[ T_{r+1} = \binom{55}{r} y^{11 - r/5} x^{r/10} \]

### Step 3: Conditions for Terms to be Free of Radical Signs
For the terms to be free of radical signs, both exponents \((11 - r/5)\) and \((r/10)\) must be integers. This leads us to the following conditions:
1. \(11 - \frac{r}{5}\) is an integer.
2. \(\frac{r}{10}\) is an integer.

### Step 4: Analyze the Conditions
From the second condition, \(\frac{r}{10}\) being an integer implies that \(r\) must be a multiple of 10. Therefore, we can write:
\[ r = 10k \quad \text{for } k = 0, 1, 2, \ldots, 5 \]
This gives us possible values for \(r\) as \(0, 10, 20, 30, 40, 50\).

### Step 5: Check the First Condition
Now, substituting \(r = 10k\) into the first condition:
\[ 11 - \frac{10k}{5} = 11 - 2k \]
This will be an integer for all integer values of \(k\).

### Step 6: Count the Valid Values of \(r\)
The values of \(k\) range from \(0\) to \(5\), giving us \(6\) possible values for \(r\):
- \(r = 0\)
- \(r = 10\)
- \(r = 20\)
- \(r = 30\)
- \(r = 40\)
- \(r = 50\)

### Conclusion
Thus, the number of terms free of radical signs in the expansion is \(6\).