If there is a term containing `x^(2r)` in `(x + (1)/(x^(2)))^(n-3)`, then

Options

A. n - 2r is a positive integral mulitple of 3
B. n - 2r is even
C. n-2r is odd
D. none of these

Answer

To find the term containing \( x^{2r} \) in the expression \( (x + \frac{1}{x^2})^{n-3} \), we will use the binomial theorem.

### Step-by-step Solution:

1. **Identify the Binomial Expansion**:
The binomial expansion of \( (a + b)^n \) is given by the general term:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
For our case, \( a = x \) and \( b = \frac{1}{x^2} \), and we have \( n = n - 3 \).

2. **Write the General Term**:
The general term in the expansion of \( (x + \frac{1}{x^2})^{n-3} \) is:
\[ T_r = \binom{n-3}{r} x^{(n-3)-r} \left(\frac{1}{x^2}\right)^r \]
Simplifying this, we get:
\[ T_r = \binom{n-3}{r} x^{(n-3)-r - 2r} = \binom{n-3}{r} x^{(n-3) - 3r} \]

3. **Set the Exponent Equal to \( 2r \)**:
We want the exponent of \( x \) to be equal to \( 2r \):
\[ n - 3 - 3r = 2r \]

4. **Rearrange the Equation**:
Rearranging the equation gives:
\[ n - 3 = 5r \]
Thus, we can express \( n \) in terms of \( r \):
\[ n = 5r + 3 \]

5. **Determine Conditions for \( n \)**:
Since \( n \) must be a positive integer, \( 5r + 3 \) must also be a positive integer. This implies \( r \) must be a non-negative integer (i.e., \( r \geq 0 \)).

6. **Conclusion**:
The term \( x^{2r} \) exists in the expansion if \( n - 3 \) is a positive integral multiple of 5. Therefore, we conclude that \( n - 2r \) must be a positive integral multiple of 5.