The sum of the rational terms in the binomial expansion of `(2^((1)/(2)) + 3^((1)/(5)))^(10)` is `:`

Options

A. 25
B. 32
C. 9
D. 41

Answer

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

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

### Step 2: Determine when the terms are rational
For \(T_r\) to be rational, both exponents \(\frac{10 - r}{2}\) and \(\frac{r}{5}\) must be integers. This means:
1. \(\frac{10 - r}{2}\) is an integer \(\Rightarrow 10 - r\) must be even \(\Rightarrow r\) must be even.
2. \(\frac{r}{5}\) is an integer \(\Rightarrow r\) must be a multiple of 5.

### Step 3: Find suitable values of \(r\)
The even numbers between 0 and 10 that are multiples of 5 are:
- \(r = 0\)
- \(r = 10\)

### Step 4: Calculate the rational terms for these values of \(r\)
1. For \(r = 0\):
\[ T_0 = \binom{10}{0} 2^{\frac{10-0}{2}} 3^{\frac{0}{5}} = 1 \cdot 2^{5} \cdot 1 = 32 \]

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

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

Thus, the sum of the rational terms in the binomial expansion of \((2^{\frac{1}{2}} + 3^{\frac{1}{5}})^{10}\) is \(\boxed{41}\).

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