Question

The independent term in the expansion of \( (1 + x + 2x^2) \left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9 \) is:

• A. \( \frac{18}{7} \)
• B. \( \frac{7}{18} \)
• C. \( \frac{-7}{18} \)
• D. \( \frac{-18}{7} \)

Hint:

When finding the independent term in a binomial expansion, carefully consider the powers of \( x \) and look for terms that cancel out.

Answer 1

The Correct Option is B

To find the independent term (constant term) in the expansion of: \[ (1 + x + 2x^2) \left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9, \] we proceed as follows: 
Step 1: Expand \( \left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9 \) Using the binomial theorem, the general term in the expansion of \( \left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9 \) is: \[ T_k = \binom{9}{k} \left( \frac{3x^2}{2} \right)^{9-k} \left( -\frac{1}{3x} \right)^k. \] Simplify the exponents of \( x \): \[ T_k = \binom{9}{k} \left( \frac{3}{2} \right)^{9-k} \left( -\frac{1}{3} \right)^k x^{2(9-k)} x^{-k}. \] Combine the exponents of \( x \): \[ T_k = \binom{9}{k} \left( \frac{3}{2} \right)^{9-k} \left( -\frac{1}{3} \right)^k x^{18 - 3k}. \] Step 2: Multiply by \( (1 + x + 2x^2) \) The expression becomes: \[ (1 + x + 2x^2) \cdot T_k. \] The exponents of \( x \) in the product are: \[ 18 - 3k, \quad 19 - 3k, \quad 20 - 3k. \] We need to find the value of \( k \) such that one of these exponents is 0 (to get the independent term). Step 3: Solve for \( k \) 1. For \( 18 - 3k = 0 \): \[ 18 - 3k = 0 \quad \Rightarrow \quad k = 6. \] 2. For \( 19 - 3k = 0 \): \[ 19 - 3k = 0 \quad \Rightarrow \quad k = \frac{19}{3}. \] This is not an integer, so it is invalid. 3. For \( 20 - 3k = 0 \): \[ 20 - 3k = 0 \quad \Rightarrow \quad k = \frac{20}{3}. \] This is not an integer, so it is invalid. Thus, the only valid case is \( k = 6 \). Step 4: Compute the term for \( k = 6 \) Substitute \( k = 6 \) into \( T_k \): \[ T_6 = \binom{9}{6} \left( \frac{3}{2} \right)^{3} \left( -\frac{1}{3} \right)^6 x^{18 - 18}. \] Simplify: \[ T_6 = \binom{9}{6} \left( \frac{27}{8} \right) \left( \frac{1}{729} \right). \] Calculate \( \binom{9}{6} \): \[ \binom{9}{6} = \binom{9}{3} = 84. \] Thus: \[ T_6 = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} = 84 \cdot \frac{27}{5832} = 84 \cdot \frac{1}{216} = \frac{84}{216} = \frac{7}{18}. \] Step 5: Multiply by \( (1 + x + 2x^2) \) Since \( k = 6 \) corresponds to the exponent \( x^0 \), the independent term is: \[ 1 \cdot T_6 = \frac{7}{18}. \] Final Answer: \[ \boxed{\frac{7}{18}} \]