The sum of the rational terms in expansion of `(sqrt2+3^(1//5))^10` is

Options

A. 41
B. 42
C. 32
D. 38

Answer

To find the sum of the rational terms in the expansion of \((\sqrt{2} + 3^{1/5})^{10}\), we will 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:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
For our expression, \(a = \sqrt{2}\), \(b = 3^{1/5}\), and \(n = 10\). Thus, the general term becomes:
\[ T_r = \binom{10}{r} (\sqrt{2})^{10-r} (3^{1/5})^r \]
This simplifies to:
\[ T_r = \binom{10}{r} (2^{1/2})^{10-r} (3^{1/5})^r = \binom{10}{r} 2^{(10-r)/2} 3^{r/5} \]

### Step 2: Determine Conditions for Rational Terms
For \(T_r\) to be a rational term, both exponents \((10 - r)/2\) and \(r/5\) must be integers. This leads to the following conditions:
1. \((10 - r)/2\) is an integer, which implies \(10 - r\) must be even. Therefore, \(r\) must be even.
2. \(r/5\) is an integer, which implies \(r\) must be a multiple of 5.

### Step 3: Find Possible Values of \(r\)
Since \(r\) must be even and a multiple of 5, the possible values of \(r\) in the range from 0 to 10 are:
- \(r = 0\)
- \(r = 10\)

### Step 4: Calculate the Rational Terms
Now we will calculate the rational terms for \(r = 0\) and \(r = 10\).

1. **For \(r = 0\)**:
\[ T_0 = \binom{10}{0} 2^{(10-0)/2} 3^{0/5} = 1 \cdot 2^{10/2} \cdot 1 = 2^5 = 32 \]

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

### Step 5: Sum of the Rational Terms
Now, we sum the rational terms:
\[ \text{Sum} = T_0 + T_{10} = 32 + 9 = 41 \]

### Final Answer
The sum of the rational terms in the expansion of \((\sqrt{2} + 3^{1/5})^{10}\) is **41**.

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