The middle term in the expansion of` ( (10)/(x) + (x)/(10) )^10` is :

Options

A. `""^7C_5`
B. `""^8C_5`
C. `""^9C_5`
D. `""^10C_5`

Answer

To find the middle term in the expansion of \( \left( \frac{10}{x} + \frac{x}{10} \right)^{10} \), we can 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, \( a = \frac{10}{x} \), \( b = \frac{x}{10} \), and \( n = 10 \).

### Step 2: Write the General Term for Our Expansion
Substituting the values into the formula, we get:
\[ T_{r+1} = \binom{10}{r} \left(\frac{10}{x}\right)^{10-r} \left(\frac{x}{10}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{10}{r} \frac{10^{10-r}}{x^{10-r}} \cdot \frac{x^r}{10^r} = \binom{10}{r} \frac{10^{10}}{10^r} \cdot \frac{x^r}{x^{10-r}} = \binom{10}{r} \frac{10^{10}}{10^r} \cdot \frac{1}{x^{10-2r}} \]
Thus, we have:
\[ T_{r+1} = \binom{10}{r} \frac{10^{10}}{10^r x^{10-2r}} \]

### Step 3: Determine the Middle Term
The total number of terms in the expansion is \( n + 1 = 10 + 1 = 11 \). The middle term is the \( 6^{th} \) term, which corresponds to \( r + 1 = 6 \) or \( r = 5 \).

### Step 4: Substitute \( r = 5 \) into the General Term
Now, we substitute \( r = 5 \) into the general term:
\[ T_{6} = \binom{10}{5} \frac{10^{10}}{10^5 x^{10-10}} = \binom{10}{5} \frac{10^{10}}{10^5} = \binom{10}{5} \cdot 10^{5} \]

### Step 5: Calculate \( \binom{10}{5} \)
The binomial coefficient \( \binom{10}{5} \) is calculated as follows:
\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Step 6: Final Calculation
Now, substituting back, we find:
\[ T_{6} = 252 \cdot 10^{5} = 25200000 \]

### Conclusion
Thus, the middle term in the expansion of \( \left( \frac{10}{x} + \frac{x}{10} \right)^{10} \) is:
\[ \boxed{25200000} \]