The coefficient of `11^(th)` term from end in the expansion of `(2x - (1)/(x^(2)))^(25)` is

Options

A. `-^(25)C_(15)`
B. `""^(25)C_(15)`
C. `-^(25)C_(16)`
D. `""^(25)C_(16)`

Answer

To find the coefficient of the 11th term from the end in the expansion of \( (2x - \frac{1}{x^2})^{25} \), we will follow these steps:

### Step 1: Identify the General Term
The general term (T_r) 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, \( a = 2x \), \( b = -\frac{1}{x^2} \), and \( n = 25 \). Therefore, the general term can be expressed as:
\[ T_{r+1} = \binom{25}{r} (2x)^{25-r} \left(-\frac{1}{x^2}\right)^r \]

### Step 2: Simplify the General Term
Now, simplifying the general term:
\[ T_{r+1} = \binom{25}{r} (2^{25-r} x^{25-r}) \left(-1\right)^r \left(\frac{1}{x^{2r}}\right) \]
This gives us:
\[ T_{r+1} = \binom{25}{r} (-1)^r 2^{25-r} x^{25-r-2r} \]
\[ T_{r+1} = \binom{25}{r} (-1)^r 2^{25-r} x^{25-3r} \]

### Step 3: Find the 11th Term from the End
The 11th term from the end corresponds to the \( (25 - 11 + 1) = 15 \)th term from the beginning. Thus, we need to find \( T_{15} \), which means \( r = 14 \):
\[ T_{15} = \binom{25}{14} (-1)^{14} 2^{25-14} x^{25-3 \cdot 14} \]
\[ T_{15} = \binom{25}{14} (2^{11}) x^{-17} \]

### Step 4: Coefficient of the 11th Term from the End
The coefficient of the 11th term from the end (which is \( T_{15} \)) is:
\[ \text{Coefficient} = \binom{25}{14} \cdot 2^{11} \]

### Step 5: Calculate the Coefficient
We can also express \( \binom{25}{14} \) as \( \binom{25}{11} \) since \( \binom{n}{r} = \binom{n}{n-r} \):
\[ \text{Coefficient} = \binom{25}{11} \cdot 2^{11} \]

### Final Answer
Thus, the coefficient of the 11th term from the end in the expansion of \( (2x - \frac{1}{x^2})^{25} \) is:
\[ \binom{25}{11} \cdot 2^{11} \]

---