In the expansion of `((x^2)/(2) - (2)/(x^2))^8`, the third term from the last is

Options

A. `(448)/(x^6)`
B. `(428)/(x^8)`
C. `(324)/(x^2)`
D. `(448)/(x^8)`

Answer

To find the third term from the last in the expansion of \(\left(\frac{x^2}{2} - \frac{2}{x^2}\right)^8\), we can follow these steps:

### Step 1: Identify the terms in the binomial expansion
The general term \(T_{r+1}\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, we have:
- \(a = \frac{x^2}{2}\)
- \(b = -\frac{2}{x^2}\)
- \(n = 8\)

### Step 2: Determine the total number of terms
The total number of terms in the expansion is \(n + 1 = 8 + 1 = 9\). Therefore, the terms are numbered from \(T_1\) to \(T_9\).

### Step 3: Find the third term from the last
The third term from the last is \(T_{9-3+1} = T_7\).

### Step 4: Calculate \(T_7\)
Using the formula for the general term:
\[ T_{7} = \binom{8}{6} \left(\frac{x^2}{2}\right)^{8-6} \left(-\frac{2}{x^2}\right)^{6} \]

### Step 5: Simplify the expression
Calculating each part:
1. \(\binom{8}{6} = \binom{8}{2} = \frac{8!}{6! \cdot 2!} = \frac{8 \times 7}{2 \times 1} = 28\)
2. \(\left(\frac{x^2}{2}\right)^{2} = \frac{x^4}{4}\)
3. \(\left(-\frac{2}{x^2}\right)^{6} = \frac{(-2)^6}{(x^2)^6} = \frac{64}{x^{12}}\)

Putting it all together:
\[ T_7 = 28 \cdot \frac{x^4}{4} \cdot \frac{64}{x^{12}} = 28 \cdot \frac{64}{4} \cdot \frac{x^4}{x^{12}} = 28 \cdot 16 \cdot \frac{1}{x^8} \]

### Step 6: Final calculation
Calculating \(28 \cdot 16\):
\[ 28 \cdot 16 = 448 \]
Thus, we have:
\[ T_7 = \frac{448}{x^8} \]

### Conclusion
The third term from the last in the expansion of \(\left(\frac{x^2}{2} - \frac{2}{x^2}\right)^8\) is:
\[ \frac{448}{x^8} \]

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