Question

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

• A. 8
• B. 9
• C. \( \frac{80}{9} \)
• D. \( \frac{29}{3} \)

Hint:

To find specific terms in a binomial expansion, use the binomial theorem to expand and then solve for the term that gives the desired power of \( x \).

Answer 1

The Correct Option is C

We are asked to find the coefficient of \( x^5 \) in the expansion of: \[ \left( 2x^3 - \frac{1}{3x^2} \right)^5. \] We will use the binomial theorem to expand this expression. The binomial expansion of \( (a + b)^n \) is given by: \[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r. \] In this case, let: \[ a = 2x^3, \quad b = -\frac{1}{3x^2}, \quad n = 5. \] Thus, the general term in the expansion is: \[ T_r = \binom{5}{r} (2x^3)^{5-r} \left(-\frac{1}{3x^2}\right)^r. \] Step 1:
Simplify the general term: \[ T_r = \binom{5}{r} 2^{5-r} x^{3(5-r)} \left( -\frac{1}{3} \right)^r x^{-2r}. \] This simplifies further to: \[ T_r = \binom{5}{r} 2^{5-r} \left( -\frac{1}{3} \right)^r x^{15 - 3r - 2r}. \] \[ T_r = \binom{5}{r} 2^{5-r} \left( -\frac{1}{3} \right)^r x^{15 - 5r}. \] Step 2:
We need the coefficient of \( x^5 \). For this to occur, the exponent of \( x \) in the general term must be 5. Thus, we set: \[ 15 - 5r = 5. \] Solving for \( r \): \[ 15 - 5r = 5 \quad \Rightarrow \quad 5r = 10 \quad \Rightarrow \quad r = 2. \] Step 3:
Substitute \( r = 2 \) into the general term to find the coefficient of \( x^5 \): \[ T_2 = \binom{5}{2} 2^{5-2} \left( -\frac{1}{3} \right)^2 x^{15 - 5(2)}. \] This simplifies to: \[ T_2 = \binom{5}{2} 2^3 \left( -\frac{1}{3} \right)^2 x^5. \] Now calculate each part: \[ \binom{5}{2} = 10, \quad 2^3 = 8, \quad \left( -\frac{1}{3} \right)^2 = \frac{1}{9}. \] Thus: \[ T_2 = 10 \times 8 \times \frac{1}{9} x^5 = \frac{80}{9} x^5. \] Step 4:
The coefficient of \( x^5 \) is \( \frac{80}{9} \). Thus, the correct answer is: \[ \boxed{\frac{80}{9}}. \]

Answer 2

To find the coefficient of \( x^5 \) in the expansion of \( \left( 2x^3 - \frac{1}{3x^2} \right)^5 \):

Step 1: Use the binomial theorem:
\[ \left(a + b\right)^5 = \sum_{k=0}^5 \binom{5}{k} a^{5-k} b^k \]
where \( a = 2x^3 \) and \( b = -\frac{1}{3x^2} \).

Step 2: Write the general term \( T_{k+1} \):
\[ T_{k+1} = \binom{5}{k} (2x^3)^{5-k} \left(-\frac{1}{3x^2}\right)^k = \binom{5}{k} 2^{5-k} (-1)^k \frac{1}{3^k} x^{3(5-k)} x^{-2k} \]

Step 3: Simplify the power of \( x \):
\[ x^{3(5-k)} \cdot x^{-2k} = x^{15 - 3k - 2k} = x^{15 - 5k} \]

Step 4: We want the power of \( x \) to be 5:
\[ 15 - 5k = 5 \implies 5k = 10 \implies k = 2 \]

Step 5: Substitute \( k = 2 \) into the term:
\[ T_3 = \binom{5}{2} 2^{3} (-1)^2 \frac{1}{3^2} x^5 = 10 \times 8 \times 1 \times \frac{1}{9} x^5 = \frac{80}{9} x^5 \]

Therefore, the coefficient of \( x^5 \) is:
\[ \boxed{\frac{80}{9}} \]