Question

The coefficient of x5 in the expansion of \((2x^3 - \frac{1}{3x^2})^5\) is

• A. 8
• B. 9
• C. \(\frac{80}{9}\)
• D. \(\frac{26}{3}\)

Answer 1

The Correct Option is C

The general term for the expansion of \( \left( 2x^3 - \frac{1}{3}x^2 \right)^5 \) is: \[ T_{r+1} = 5C_r \left( - \frac{1}{3}x^2 \right)^r \left( 2x^3 \right)^{5-r} \] \[ T_{r+1} = 5C_r \left( -1 \right)^r \left( \frac{1}{3} \right)^r x^{2r} \cdot 2^{5-r} x^{3(5-r)} \] \[ T_{r+1} = 5C_r (-1)^r \left( \frac{1}{3} \right)^r 2^{5-r} x^{2r + 3(5-r)} = 5C_r (-1)^r \left( \frac{1}{3} \right)^r 2^{5-r} x^{15 - r} \] We need to find the term where the power of \( x \) is 5. Thus: \[ 15 - r = 5 \quad \Rightarrow \quad r = 10 \] Substitute \( r = 2 \) (the required power of \( x^5 \)): \[ \text{Coefficient of } x^5 = 5C_2 (-1)^2 \left( \frac{1}{3} \right)^2 2^{3} \] \[ = 10 \times \frac{1}{9} \times 8 = \frac{80}{9} \]