The sum of the rational terms in the expansion of
`(sqrt(2)+ root(5)(3))^(10)` is

Options

A. 32
B. 9
C. 41
D. none of these

Explanation

To find the sum of the rational terms in the expansion of \( \left( \sqrt{2} + \frac{1}{\sqrt[5]{3}} \right)^{10} \), we will follow these steps:

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = \sqrt{2} \), \( b = \frac{1}{\sqrt[5]{3}} \), and \( n = 10 \).

Thus, the general term becomes:
\[ T_{r+1} = \binom{10}{r} (\sqrt{2})^{10-r} \left(\frac{1}{\sqrt[5]{3}}\right)^r \]

### Step 2: Simplify the General Term
We simplify \( T_{r+1} \):
\[ T_{r+1} = \binom{10}{r} (2^{\frac{10-r}{2}}) \left(3^{-\frac{r}{5}}\right) \]
This can be rewritten as:
\[ T_{r+1} = \binom{10}{r} 2^{\frac{10-r}{2}} 3^{-\frac{r}{5}} \]

### Step 3: Determine Conditions for Rational Terms
For \( T_{r+1} \) to be a rational term, both exponents \( \frac{10-r}{2} \) and \( -\frac{r}{5} \) must be integers.

1. \( \frac{10 - r}{2} \) is an integer implies \( 10 - r \) is even, hence \( r \) must be even.
2. \( -\frac{r}{5} \) is an integer implies \( r \) must be a multiple of 5.

### Step 4: Find Possible Values of \( r \)
Since \( r \) must be even and a multiple of 5, the possible values of \( r \) in the range \( 0 \leq r \leq 10 \) are:
- \( r = 0 \)
- \( r = 10 \)

### Step 5: Calculate the Rational Terms
Now, we calculate the rational terms for \( r = 0 \) and \( r = 10 \).

1. For \( r = 0 \):
\[ T_{1} = \binom{10}{0} 2^{\frac{10-0}{2}} 3^{-\frac{0}{5}} = 1 \cdot 2^{5} \cdot 1 = 32 \]

2. For \( r = 10 \):
\[ T_{11} = \binom{10}{10} 2^{\frac{10-10}{2}} 3^{-\frac{10}{5}} = 1 \cdot 2^{0} \cdot 3^{-2} = 1 \cdot 1 \cdot \frac{1}{9} = \frac{1}{9} \]

### Step 6: Sum of the Rational Terms
Now we sum the rational terms:
\[ \text{Sum} = T_{1} + T_{11} = 32 + \frac{1}{9} \]
To add these, convert \( 32 \) into a fraction:
\[ 32 = \frac{288}{9} \]
Thus,
\[ \text{Sum} = \frac{288}{9} + \frac{1}{9} = \frac{289}{9} \]

### Final Answer
The sum of the rational terms in the expansion is:
\[ \frac{289}{9} \]