Find the `6^(th)` term from the end in the expansion of `(2x-(1)/(x^(2)))^(10)` .

Answer

To find the 6th term from the end in the expansion of \( (2x - \frac{1}{x^2})^{10} \), we can follow these steps:

### Step 1: Identify the parameters
In the expression \( (2x - \frac{1}{x^2})^{10} \):
- \( n = 10 \) (the exponent)
- We need to find the 6th term from the end, which corresponds to the \( (n - r + 1) \)-th term from the beginning, where \( r = 6 \).

### Step 2: Calculate the term number from the beginning
The term number from the beginning can be calculated as:
\[ n - r + 1 = 10 - 6 + 1 = 5 \]
So, we need to find the 5th term in the expansion.

### Step 3: Use the binomial theorem
The general term \( T_{r+1} \) in the binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} (a)^{n-r} (b)^{r} \]
where \( a = 2x \) and \( b = -\frac{1}{x^2} \).

For the 5th term, we have:
- \( r = 4 \) (since we are looking for the 5th term, \( r \) is one less than the term number)

### Step 4: Substitute values into the formula
Now we substitute \( n = 10 \), \( r = 4 \), \( a = 2x \), and \( b = -\frac{1}{x^2} \):
\[ T_5 = \binom{10}{4} (2x)^{10-4} \left(-\frac{1}{x^2}\right)^{4} \]

### Step 5: Calculate the binomial coefficient
Calculate \( \binom{10}{4} \):
\[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]

### Step 6: Calculate the powers
Now calculate \( (2x)^6 \) and \( \left(-\frac{1}{x^2}\right)^4 \):
\[ (2x)^6 = 2^6 x^6 = 64x^6 \]
\[ \left(-\frac{1}{x^2}\right)^4 = \frac{1}{x^8} \]

### Step 7: Combine the results
Now substitute back into the term:
\[ T_5 = 210 \cdot 64x^6 \cdot \frac{1}{x^8} \]
\[ T_5 = 210 \cdot 64 \cdot x^{6-8} = 210 \cdot 64 \cdot x^{-2} \]

### Step 8: Calculate the final coefficient
Calculate \( 210 \cdot 64 \):
\[ 210 \cdot 64 = 13440 \]

### Final Result
Thus, the 5th term (which is the 6th term from the end) is:
\[ T_5 = 13440 x^{-2} \]

### Conclusion
The required 6th term from the end in the expansion of \( (2x - \frac{1}{x^2})^{10} \) is:
\[ \boxed{13440 x^{-2}} \]