The sum of the binomial coefficients in the expansion of `((2x)/(3) + (3)/(2x^2))^n` is 64 then the term independent of x is

Options

A. `20//3`
B. `3//20`
C. `10//3`
D. `3//10`

Answer

To solve the problem, we need to find the term independent of \( x \) in the expansion of

\[ \left(\frac{2x}{3} + \frac{3}{2x^2}\right)^n \]

given that the sum of the binomial coefficients in this expansion is 64.

### Step-by-step Solution:

1. **Understanding the Sum of Binomial Coefficients**:
The sum of the binomial coefficients in the expansion of \( (a + b)^n \) is given by \( 2^n \). Here, we have:
\[ \left(\frac{2x}{3} + \frac{3}{2x^2}\right)^n \]
Therefore, the sum of the coefficients is:
\[ 2^n = 64 \]
Since \( 64 = 2^6 \), we can equate:
\[ n = 6 \]

**Hint**: Remember that the sum of coefficients in a binomial expansion is \( 2^n \).

2. **Finding the General Term**:
The general term \( T_{r+1} \) in the expansion can be expressed as:
\[ T_{r+1} = \binom{n}{r} \left(\frac{2x}{3}\right)^{n-r} \left(\frac{3}{2x^2}\right)^r \]
Substituting \( n = 6 \):
\[ T_{r+1} = \binom{6}{r} \left(\frac{2x}{3}\right)^{6-r} \left(\frac{3}{2x^2}\right)^r \]

3. **Simplifying the General Term**:
Simplifying the term:
\[ T_{r+1} = \binom{6}{r} \left(\frac{2^{6-r} x^{6-r}}{3^{6-r}}\right) \left(\frac{3^r}{(2^r)(x^{2r})}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{6}{r} \frac{2^{6-r} \cdot 3^r}{3^{6-r} \cdot 2^r} x^{6 - r - 2r} \]
\[ = \binom{6}{r} \frac{2^{6-r} \cdot 3^r}{3^{6-r} \cdot 2^r} x^{6 - 3r} \]

4. **Finding the Term Independent of \( x \)**:
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ 6 - 3r = 0 \implies 6 = 3r \implies r = 2 \]

**Hint**: Set the exponent of \( x \) to zero to find \( r \).

5. **Finding the Required Term**:
Substitute \( r = 2 \) into the general term:
\[ T_{3} = \binom{6}{2} \frac{2^{6-2} \cdot 3^2}{3^{6-2} \cdot 2^2} \]
\[ = \binom{6}{2} \frac{2^4 \cdot 9}{81 \cdot 4} \]
\[ = 15 \cdot \frac{16 \cdot 9}{81 \cdot 4} \]
\[ = 15 \cdot \frac{144}{324} \]
\[ = 15 \cdot \frac{4}{9} = \frac{60}{9} = \frac{20}{3} \]

6. **Final Answer**:
The term independent of \( x \) is:
\[ \frac{20}{3} \]