In the expansion of `(root(5)(3) + root(7)(2))^24` , then rational term is

Options

A. `T_14`
B. `T_16`
C. `T_15 `
D. `T_7`

Answer

To find the rational term in the expansion of \((\sqrt[5]{3} + \sqrt[7]{2})^{24}\), we can follow these steps:

### Step 1: Write 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 case, \(a = \sqrt[5]{3}\), \(b = \sqrt[7]{2}\), and \(n = 24\). Thus, the general term becomes:
\[ T_r = \binom{24}{r} \left(\sqrt[5]{3}\right)^{24-r} \left(\sqrt[7]{2}\right)^r \]
This can be rewritten as:
\[ T_r = \binom{24}{r} 3^{\frac{24-r}{5}} 2^{\frac{r}{7}} \]

### Step 2: Conditions for Rational Terms
For \(T_r\) to be a rational term, both exponents \(\frac{24 - r}{5}\) and \(\frac{r}{7}\) must be integers. This leads to the following conditions:
1. \(\frac{24 - r}{5}\) is an integer, which implies \(24 - r \equiv 0 \mod 5\).
2. \(\frac{r}{7}\) is an integer, which implies \(r \equiv 0 \mod 7\).

### Step 3: Solve the Conditions
From the first condition, we can express it as:
\[ 24 - r = 5k \quad \text{for some integer } k \]
This implies:
\[ r = 24 - 5k \]

From the second condition, we have:
\[ r = 7m \quad \text{for some integer } m \]

### Step 4: Equate and Find Possible Values of \(r\)
Substituting \(r = 7m\) into \(r = 24 - 5k\):
\[ 7m = 24 - 5k \]
Rearranging gives:
\[ 5k + 7m = 24 \]

### Step 5: Find Integer Solutions
We need to find integer pairs \((k, m)\) that satisfy this equation. We can test integer values for \(m\):
- For \(m = 0\): \(5k = 24 \Rightarrow k = 4.8\) (not an integer)
- For \(m = 1\): \(5k + 7 = 24 \Rightarrow 5k = 17 \Rightarrow k = 3.4\) (not an integer)
- For \(m = 2\): \(5k + 14 = 24 \Rightarrow 5k = 10 \Rightarrow k = 2\) (integer)
- For \(m = 3\): \(5k + 21 = 24 \Rightarrow 5k = 3 \Rightarrow k = 0.6\) (not an integer)

The only integer solution we found is \(m = 2\) and \(k = 2\):
\[ r = 7m = 7 \times 2 = 14 \]

### Step 6: Find the Term Number
The term corresponding to \(r = 14\) is:
\[ T_{14} = \binom{24}{14} \left(\sqrt[5]{3}\right)^{10} \left(\sqrt[7]{2}\right)^{14} \]
The term number is \(r + 1 = 14 + 1 = 15\).

### Final Answer
Thus, the rational term in the expansion is the 15th term.