If `(r+1)^(th)` term in the expasnion of `(a^(3)/3-2/a^(2))^(10)` contains ` a^(20)` then the value of `r` is equal to

Options

A. 3
B. 2
C. 4
D. 1

Answer

To solve the problem, we need to find the value of \( r \) such that the \( (r+1)^{th} \) term in the expansion of \( \left( \frac{a^3}{3} - \frac{2}{a^2} \right)^{10} \) contains \( a^{20} \).

### Step-by-Step Solution:

1. **Identify the general term in the binomial expansion**:
The general term \( T_{r+1} \) in the expansion of \( (x + y)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]
For our expression, let \( x = \frac{a^3}{3} \) and \( y = -\frac{2}{a^2} \), and \( n = 10 \).

2. **Write the specific term**:
The \( (r+1)^{th} \) term can be expressed as:
\[ T_{r+1} = \binom{10}{r} \left(\frac{a^3}{3}\right)^{10-r} \left(-\frac{2}{a^2}\right)^r \]

3. **Simplify the term**:
Now, we simplify this expression:
\[ T_{r+1} = \binom{10}{r} \left(\frac{a^3}{3}\right)^{10-r} \left(-2\right)^r \left(a^{-2}\right)^r \]
This becomes:
\[ T_{r+1} = \binom{10}{r} \frac{(-2)^r}{3^{10-r}} a^{3(10-r) - 2r} \]

4. **Combine the powers of \( a \)**:
The exponent of \( a \) in \( T_{r+1} \) is:
\[ 3(10 - r) - 2r = 30 - 3r - 2r = 30 - 5r \]

5. **Set the exponent equal to 20**:
We need this exponent to equal 20:
\[ 30 - 5r = 20 \]

6. **Solve for \( r \)**:
Rearranging gives:
\[ 30 - 20 = 5r \implies 10 = 5r \implies r = 2 \]

### Conclusion:
Thus, the value of \( r \) is \( 2 \).