The term independent of x in the expansion of `(1- x^(2) + 3x^(3)) ((5)/(2) x^(3)- (1)/(5x^(2)))^(11), x ne 0` is:

Options

A. `(7)/(40)`
B. `(33)/(200)`
C. `(39)/(200)`
D. `(11)/(50)`

Answer

To find the term independent of \( x \) in the expansion of \( (1 - x^2 + 3x^3) \left( \frac{5}{2} x^3 - \frac{1}{5x^2} \right)^{11} \), we will follow these steps:

### Step 1: Identify the Binomial Expansion
The expression can be rewritten as:
\[ (1 - x^2 + 3x^3) \left( \frac{5}{2} x^3 - \frac{1}{5x^2} \right)^{11} \]
The second part of the expression is a binomial expansion.

### Step 2: General Term of the Binomial Expansion
The general term \( T_r \) in the expansion of \( \left( a + b \right)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = \frac{5}{2} x^3 \), \( b = -\frac{1}{5x^2} \), and \( n = 11 \).

Thus, the general term \( T_r \) becomes:
\[ T_r = \binom{11}{r} \left( \frac{5}{2} x^3 \right)^{11-r} \left( -\frac{1}{5x^2} \right)^{r} \]

### Step 3: Simplify the General Term
Now, simplifying \( T_r \):
\[ T_r = \binom{11}{r} \left( \frac{5}{2} \right)^{11-r} \left( -\frac{1}{5} \right)^{r} x^{3(11-r)} x^{-2r} \]
This simplifies to:
\[ T_r = \binom{11}{r} \left( \frac{5}{2} \right)^{11-r} \left( -1 \right)^{r} \frac{1}{5^r} x^{33 - 5r} \]

### Step 4: Find the Condition for the Term Independent of \( x \)
To find the term independent of \( x \), we set the exponent of \( x \) to zero:
\[ 33 - 5r = 0 \implies 5r = 33 \implies r = \frac{33}{5} \]
Since \( r \) must be an integer, we check for integer values around \( \frac{33}{5} = 6.6 \).

### Step 5: Check Integer Values of \( r \)
1. For \( r = 6 \):
\[ 33 - 5(6) = 33 - 30 = 3 \quad (\text{not independent}) \]
2. For \( r = 7 \):
\[ 33 - 5(7) = 33 - 35 = -2 \quad (\text{not independent}) \]

### Step 6: Calculate the Coefficient for \( r = 6 \) and \( r = 7 \)
Since neither \( r = 6 \) nor \( r = 7 \) gives a term independent of \( x \), we will check the coefficients for \( r = 6 \) and \( r = 7 \).

- For \( r = 6 \):
\[ T_6 = \binom{11}{6} \left( \frac{5}{2} \right)^{5} \left( -1 \right)^{6} \frac{1}{5^6} x^{3} \]

- For \( r = 7 \):
\[ T_7 = \binom{11}{7} \left( \frac{5}{2} \right)^{4} \left( -1 \right)^{7} \frac{1}{5^7} x^{-2} \]

### Step 7: Combine with the First Part
Now we will multiply these with the first part \( (1 - x^2 + 3x^3) \) to find the independent term.

### Step 8: Final Calculation
After calculating the coefficients from \( T_6 \) and \( T_7 \) and combining them with \( (1 - x^2 + 3x^3) \), we will find the final coefficient of the term independent of \( x \).