Rewritten Solution: Finding the Coefficient of a Specific Term in a Binomial Expansion

---

1. Understanding the Binomial Expansion and General Term

The problem asks for the coefficient of a specific power of $x$ in a binomial expansion. The key to solving such problems lies in the Binomial Theorem, which provides a formula for expanding expressions of the form $(a+b)^n$.

The general term, also known as the $(r+1)^{th}$ term, in the expansion of $(a+b)^n$ is given by: $T_{r+1} = {}^{n}C_r \, a^{n-r} \, b^r$ Where:

• $T_{r+1}$ represents the $(r+1)^{th}$ term.
• $n$ is the power to which the binomial is raised (the exponent of the entire expression).
• $r$ is an integer ranging from $0$ to $n$, which determines the specific term.
• ${}^{n}C_r$ (read as "n choose r") is the binomial coefficient, calculated as $\frac{n!}{r!(n-r)!}$.
• $a$ is the first term of the binomial.
• $b$ is the second term of the binomial.

Our strategy is to first write down the general term for the given expression, isolate the powers of $x$, find the value of $r$ that yields the desired power of $x$, and then substitute that $r$ back into the numerical part of the general term to find the coefficient.

---

2. Applying the General Term to the Given Expression

The given expression is $\left(\frac{4x}{5} + \frac{5}{2x^2}\right)^9$. Comparing this to $(a+b)^n$, we can identify:

• $a = \frac{4x}{5}$
• $b = \frac{5}{2x^2}$
• $n = 9$

Now, substitute these into the general term formula: $T_{r+1} = {}^{9}C_r \left(\frac{4x}{5}\right)^{9-r} \left(\frac{5}{2x^2}\right)^r$

To simplify and determine the power of $x$, we separate the numerical coefficients from the variable parts: $T_{r+1} = {}^{9}C_r \left(\frac{4}{5}\right)^{9-r} (x)^{9-r} \left(\frac{5}{2}\right)^r \left(\frac{1}{x^2}\right)^r$ $T_{r+1} = {}^{9}C_r \left(\frac{4}{5}\right)^{9-r} \left(\frac{5}{2}\right)^r \cdot x^{9-r} \cdot x^{-2r}$ $T_{r+1} = {}^{9}C_r \left(\frac{4}{5}\right)^{9-r} \left(\frac{5}{2}\right)^r \cdot x^{9-r-2r}$ $T_{r+1} = {}^{9}C_r \left(\frac{4}{5}\right)^{9-r} \left(\frac{5}{2}\right)^r \cdot x^{9-3r}$

This is the fully simplified general term, showing both its numerical coefficient part and its variable part.

---

3. Finding the Value of 'r' for the Desired Term

We are looking for the coefficient of $x^{-6}$. From the simplified general term, the exponent of $x$ is $9-3r$. Therefore, to find the term containing $x^{-6}$, we must equate the exponent of $x$ to $-6$: $9 - 3r = -6$ Now, we solve this linear equation for $r$: $-3r = -6 - 9$ $-3r = -15$ $r = \frac{-15}{-3}$ $r = 5$

Tip: The value of $r$ must always be a non-negative integer ($0 \le r \le n$). If you obtain a fractional or negative value for $r$, it means that the specified power of $x$ does not exist in the expansion. In this case, $r=5$ is a valid integer between $0$ and $9$.

---

4. Calculating the Coefficient

Now that we have found $r=5$, we substitute this value back into the numerical part of our general term, which is everything except $x^{9-3r}$: $\text{Coefficient of } x^{-6} = {}^{9}C_5 \left(\frac{4}{5}\right)^{9-5} \left(\frac{5}{2}\right)^5$ $\text{Coefficient of } x^{-6} = {}^{9}C_5 \left(\frac{4}{5}\right)^4 \left(\frac{5}{2}\right)^5$

Let's calculate each part:

• Binomial Coefficient ${}^{9}C_5$: Using the identity ${}^{n}C_r = {}^{n}C_{n-r}$, we have ${}^{9}C_5 = {}^{9}C_{9-5} = {}^{9}C_4$. ${}^{9}C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{9 \times (2 \times 4) \times 7 \times (2 \times 3)}{(4 \times 3 \times 2 \times 1)} = 9 \times 2 \times 7 = 126$

• Powers of the numerical terms: $\left(\frac{4}{5}\right)^4 = \frac{4^4}{5^4} = \frac{256}{625}$ $\left(\frac{5}{2}\right)^5 = \frac{5^5}{2^5} = \frac{3125}{32}$

Now, multiply these values together: $\text{Coefficient} = 126 \times \frac{256}{625} \times \frac{3125}{32}$ To simplify the multiplication, look for common factors between numerators and denominators:

• $3125$ is $5^5$, and $625$ is $5^4$. So, $\frac{3125}{625} = 5$.
• $256$ is $2^8$, and $32$ is $2^5$. So, $\frac{256}{32} = 8$.

Substitute these simplified terms back: $\text{Coefficient} = 126 \times \frac{256}{32} \times \frac{3125}{625}$ $\text{Coefficient} = 126 \times 8 \times 5$ $\text{Coefficient} = 126 \times 40$ $\text{Coefficient} = 5040$

Common Mistake: A frequent error is to forget to include the numerical coefficients from $a$ and $b$ when calculating the final coefficient. Ensure all parts of the numerical term (the binomial coefficient and the powers of $a$ and $b$) are correctly calculated and multiplied.

---

5. Summary and Key Takeaways

To find the coefficient of a specific power of $x$ in a binomial expansion:

1. Write down the general term, $T_{r+1}$, using the formula ${}^{n}C_r \, a^{n-r} \, b^r$.
2. Simplify the general term, collecting all powers of $x$ together.
3. Equate the resulting exponent of $x$ to the desired power (in this case, $-6$) and solve for $r$. Remember that $r$ must be a non-negative integer.
4. Substitute the value of $r$ back into the numerical part of the general term (excluding the $x$ variable) and perform the calculation carefully.

The coefficient of $x^{-6}$ in the expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x^{2}}\right)^{9}$ is $\mathbf{5040}$.