The number of rational terms in the expansion of `((25)^(1/3) + 1/(25)^(1/3))^(20)` is

Options

A. 2
B. 7
C. 6
D. 19

Answer

To find the number of rational terms in the expansion of \(((25)^{1/3} + \frac{1}{(25)^{1/3}})^{20}\), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
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, \(a = (25)^{1/3}\), \(b = \frac{1}{(25)^{1/3}}\), and \(n = 20\). Thus, we have:
\[ T_{r+1} = \binom{20}{r} \left(25^{1/3}\right)^{20-r} \left(\frac{1}{25^{1/3}}\right)^{r} \]

### Step 2: Simplify the general term
Now we simplify the general term:
\[ T_{r+1} = \binom{20}{r} \left(25^{(20-r)/3}\right) \left(25^{-r/3}\right) \]
Combining the powers of 25, we get:
\[ T_{r+1} = \binom{20}{r} \cdot 25^{\frac{20 - 2r}{3}} \]

### Step 3: Determine when the term is rational
For \(T_{r+1}\) to be a rational term, the exponent of 25 must be an integer. Therefore, we need:
\[ \frac{20 - 2r}{3} \text{ to be an integer} \]
This implies that \(20 - 2r\) must be divisible by 3.

### Step 4: Set up the divisibility condition
We can express this condition as:
\[ 20 - 2r \equiv 0 \mod 3 \]
Calculating \(20 \mod 3\):
\[ 20 \equiv 2 \mod 3 \]
Thus, we have:
\[ 2 - 2r \equiv 0 \mod 3 \implies 2r \equiv 2 \mod 3 \implies r \equiv 1 \mod 3 \]

### Step 5: Find the values of \(r\)
The values of \(r\) that satisfy \(r \equiv 1 \mod 3\) within the range \(0 \leq r \leq 20\) are:
- \(r = 1\)
- \(r = 4\)
- \(r = 7\)
- \(r = 10\)
- \(r = 13\)
- \(r = 16\)
- \(r = 19\)

### Step 6: Count the rational terms
Counting these values, we find that there are a total of 7 values of \(r\) that yield rational terms.

### Final Answer
Thus, the number of rational terms in the expansion is **7**.

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