To find the term containing $x^2$ in the expansion of $(3x - \frac{1}{2x})^8$, we'll again use the Binomial Theorem.

Step 1: Binomial Theorem Formula

The Binomial Theorem states that: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ In this case:

• $a = 3x$
• $b = -\frac{1}{2x}$
• $n = 8$

Step 2: General Term in the Expansion

The general term in the expansion is given by: $T_{k+1} = \binom{8}{k} (3x)^{8-k} \left(-\frac{1}{2x}\right)^k$

Step 3: Simplify the General Term

Simplifying $T_{k+1}$: $T_{k+1} = \binom{8}{k} \cdot (3x)^{8-k} \cdot \left(-\frac{1}{2x}\right)^k$ $T_{k+1} = \binom{8}{k} \cdot 3^{8-k} \cdot x^{8-k} \cdot \frac{(-1)^k}{2^k} \cdot x^{-k}$ $T_{k+1} = \binom{8}{k} \cdot \frac{3^{8-k} \cdot (-1)^k}{2^k} \cdot x^{8-k-k}$ $T_{k+1} = \binom{8}{k} \cdot \frac{3^{8-k} \cdot (-1)^k}{2^k} \cdot x^{8-2k}$

Step 4: Find the Value of $k$ for $x^2$

To find the term containing $x^2$, set the exponent of $x$ equal to 2: $8 - 2k = 2$ Solve for $k$: $8 - 2k = 2$ $2k = 6$ $k = 3$

Step 5: Substitute $k = 3$ into the General Term

Now substitute $k = 3$ back into the general term: $T_{4} = \binom{8}{3} \cdot \frac{3^{8-3} \cdot (-1)^3}{2^3} \cdot x^2$ $T_{4} = \binom{8}{3} \cdot \frac{3^5 \cdot (-1)}{8} \cdot x^2$ $T_{4} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \cdot \frac{243 \cdot (-1)}{8} \cdot x^2$ $T_{4} = 56 \cdot \frac{-243}{8} \cdot x^2$ $T_{4} = -1701x^2$

Final Answer

The term containing $x^2$ in the expansion of $(3x - \frac{1}{2x})^8$ is: $-1701x^2$

Would you like any more details or have any questions?

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Related Questions:

1. How do you find the coefficient of a specific term in a binomial expansion?
2. What is the term containing $x^4$ in this expansion?
3. How would the expansion change if the exponent were a different value?
4. What are the first and last terms in the expansion?
5. Can you identify a pattern in the coefficients of terms in the binomial expansion?

Tip: When dealing with binomial expansions that involve fractions or negative terms, pay close attention to how these factors influence the coefficients and the signs of each term.