Find the term independent of x (that is the constant term) in the expansion of `((sqrt(x))/(3)+(3)/(2x^(2)))^(10)`

Answer

To find the term independent of \( x \) (the constant term) in the expansion of

\[ \left( \frac{\sqrt{x}}{3} + \frac{3}{2x^2} \right)^{10}, \]

we can use the Binomial Theorem. The general term in the expansion of \( (a + b)^n \) is given by:

\[ T_r = \binom{n}{r} a^{n-r} b^r, \]

where \( n \) is the exponent, \( a \) and \( b \) are the two terms being expanded, and \( r \) is the term number starting from 0.

### Step 1: Identify \( a \) and \( b \)
In our case, we have:
- \( a = \frac{\sqrt{x}}{3} = \frac{x^{1/2}}{3} \)
- \( b = \frac{3}{2x^2} = \frac{3}{2} x^{-2} \)
- \( n = 10 \)

### Step 2: Write the general term
The general term \( T_r \) in the expansion is:

\[ T_r = \binom{10}{r} \left( \frac{x^{1/2}}{3} \right)^{10-r} \left( \frac{3}{2} x^{-2} \right)^r. \]

### Step 3: Simplify the general term
Now, we simplify \( T_r \):

\[ T_r = \binom{10}{r} \left( \frac{1}{3^{10-r}} \right) x^{\frac{10-r}{2}} \left( \frac{3^r}{2^r} \right) x^{-2r}. \]

Combining the powers of \( x \):

\[ T_r = \binom{10}{r} \frac{3^r}{3^{10-r} \cdot 2^r} x^{\frac{10-r}{2} - 2r}. \]

### Step 4: Find the exponent of \( x \)
We need the exponent of \( x \) to be 0 for the term to be constant:

\[ \frac{10 - r}{2} - 2r = 0. \]

### Step 5: Solve for \( r \)
Multiplying through by 2 to eliminate the fraction:

\[ 10 - r - 4r = 0 \implies 10 - 5r = 0 \implies 5r = 10 \implies r = 2. \]

### Step 6: Substitute \( r \) back into the general term
Now we substitute \( r = 2 \) back into the general term \( T_r \):

\[ T_2 = \binom{10}{2} \frac{3^2}{3^{10-2} \cdot 2^2} x^{0}. \]

### Step 7: Calculate the constant term
Calculating \( T_2 \):

\[ T_2 = \binom{10}{2} \frac{9}{3^8 \cdot 4}. \]

Calculating \( \binom{10}{2} \):

\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45. \]

Now substituting this back:

\[ T_2 = 45 \cdot \frac{9}{6561 \cdot 4} = \frac{405}{26244}. \]

### Final Result
Thus, the constant term in the expansion is:

\[ \frac{405}{26244}. \]