Find the number of rational terms and also find
the sum of rational terms in ` (sqrt(2) + root(3)(3) + root(6)(5) )^(10)`

Options

A. 12632
B. 1260
C. 126
D. None of these

Explanation

To solve the problem of finding the number of rational terms and the sum of rational terms in the expression \( ( \sqrt{2} + \sqrt[3]{3} + \sqrt[6]{5} )^{10} \), we will follow these steps:

### Step 1: Rewrite the expression
We can express the terms in a simpler form:
\[ \sqrt{2} = 2^{1/2}, \quad \sqrt[3]{3} = 3^{1/3}, \quad \sqrt[6]{5} = 5^{1/6} \]
Thus, we can rewrite the expression as:
\[ (2^{1/2} + 3^{1/3} + 5^{1/6})^{10} \]

### Step 2: Use the multinomial expansion
Using the multinomial theorem, the general term in the expansion of \( (x_1 + x_2 + x_3)^n \) is given by:
\[ \frac{n!}{k_1! k_2! k_3!} x_1^{k_1} x_2^{k_2} x_3^{k_3} \]
where \( k_1 + k_2 + k_3 = n \).

In our case, we have:
\[ x_1 = 2^{1/2}, \quad x_2 = 3^{1/3}, \quad x_3 = 5^{1/6}, \quad n = 10 \]
Thus, the general term becomes:
\[ T = \frac{10!}{\alpha! \beta! \gamma!} (2^{1/2})^{\alpha} (3^{1/3})^{\beta} (5^{1/6})^{\gamma} \]
where \( \alpha + \beta + \gamma = 10 \).

### Step 3: Determine conditions for rational terms
For \( T \) to be rational, the exponents of \( 2 \), \( 3 \), and \( 5 \) must be integers. This leads to the following conditions:
- The exponent of \( 2 \): \( \frac{\alpha}{2} \) must be an integer, so \( \alpha \) must be even.
- The exponent of \( 3 \): \( \frac{\beta}{3} \) must be an integer, so \( \beta \) must be a multiple of 3.
- The exponent of \( 5 \): \( \frac{\gamma}{6} \) must be an integer, so \( \gamma \) must be a multiple of 6.

### Step 4: Find possible values for \( \alpha, \beta, \gamma \)
Given \( \alpha + \beta + \gamma = 10 \):
- Possible values for \( \alpha \) (even): \( 0, 2, 4, 6, 8, 10 \)
- Possible values for \( \beta \) (multiples of 3): \( 0, 3, 6, 9 \)
- Possible values for \( \gamma \) (multiples of 6): \( 0, 6 \)

### Step 5: List valid combinations
We need to find combinations of \( (\alpha, \beta, \gamma) \) that satisfy \( \alpha + \beta + \gamma = 10 \):
1. \( (4, 0, 6) \)
2. \( (4, 6, 0) \)
3. \( (10, 0, 0) \)

### Step 6: Count the number of rational terms
From the combinations above, we find that there are 3 rational terms.

### Step 7: Calculate the sum of rational terms
Now we calculate the sum of the rational terms:
1. For \( (4, 0, 6) \):
\[ T_1 = \frac{10!}{4!0!6!} (2^{1/2})^4 (3^{1/3})^0 (5^{1/6})^6 = \frac{10!}{4!6!} \cdot 2^2 \cdot 5 = 4200 \]

2. For \( (4, 6, 0) \):
\[ T_2 = \frac{10!}{4!6!} (2^{1/2})^4 (3^{1/3})^6 (5^{1/6})^0 = \frac{10!}{4!6!} \cdot 2^2 \cdot 3^2 = 7560 \]

3. For \( (10, 0, 0) \):
\[ T_3 = \frac{10!}{10!0!0!} (2^{1/2})^{10} (3^{1/3})^0 (5^{1/6})^0 = 2^5 = 32 \]

### Step 8: Sum of rational terms
The total sum of the rational terms is:
\[ 4200 + 7560 + 32 = 11792 \]

### Final Answer
- Number of rational terms: **3**
- Sum of rational terms: **11792**