To find the coefficient of $x^5$ in the expansion of $\left(x + \frac{1}{2x}\right)^9$, we can use the Binomial Theorem:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

In this case:

• $a = x$
• $b = \frac{1}{2x}$
• $n = 9$

So the general term in the expansion is:

$T_k = \binom{9}{k} \cdot x^{9-k} \cdot \left(\frac{1}{2x}\right)^k$

Simplifying $T_k$:

$T_k = \binom{9}{k} \cdot x^{9-k} \cdot \frac{1}{(2x)^k} = \binom{9}{k} \cdot \frac{x^{9-k}}{2^k \cdot x^k} = \binom{9}{k} \cdot \frac{x^{9 - 2k}}{2^k}$

Now, we need the coefficient of $x^5$, so we set up the equation:

$9 - 2k = 5$

Solving for $k$:

$2k = 4 \Rightarrow k = 2$

Now, substitute $k = 2$ back into $T_k$ to find the coefficient:

$T_2 = \binom{9}{2} \cdot \frac{x^{9 - 2 \cdot 2}}{2^2} = \binom{9}{2} \cdot \frac{x^5}{4}$

Calculating $\binom{9}{2}$:

$\binom{9}{2} = \frac{9 \cdot 8}{2 \cdot 1} = 36$

Thus,

$T_2 = 36 \cdot \frac{x^5}{4} = 9x^5$

So, the coefficient of $x^5$ in the expansion of $\left(x + \frac{1}{2x}\right)^9$ is 9.

Would you like further clarification on this solution?

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1. What is the Binomial Theorem and how is it applied in different contexts?
2. How would the coefficient change if the power were $10$ instead of $9$?
3. What role does the term $k$ play in identifying specific terms in expansions?
4. How would you solve this if the expression was $(x + 1/x)^9$ instead?
5. Can we solve this problem by other methods like generating functions?

Tip: Understanding the general term in binomial expansions helps in identifying specific terms more efficiently.