Question

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

Hint:

In binomial expansions, identify the power of \( x \) in the general term and solve for the value of \( r \) that gives the desired exponent. Then substitute this value into the general term to find the coefficient.

Answer 1

Correct Answer: 5040

To find the coefficient of \(x^{-6}\) in the expansion of \(\left(\frac{4x}{5}+\frac{5}{2x^2}\right)^9\), we will use the binomial theorem, which states that \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).

In this problem, we have \(a = \frac{4x}{5}\) and \(b = \frac{5}{2x^2}\) with \(n = 9\). The general term in the expansion is given by:

\(T_k = \binom{9}{k} \left(\frac{4x}{5}\right)^{9-k} \left(\frac{5}{2x^2}\right)^k\).

Let's simplify this term:

• The coefficient part is \(\binom{9}{k} \left(\frac{4}{5}\right)^{9-k} \left(\frac{5}{2}\right)^k\).
• The \(x\) part is \(x^{9-k} \cdot x^{-2k} = x^{9-k-2k} = x^{9-3k}\).

We are looking for the term where the power of \(x\) is \(-6\). So, we set the exponent of \(x\) equal to \(-6\):

\(9-3k = -6\).

Solving for \(k\):

1. \(9 - 3k = -6\)
2. \(9 + 6 = 3k\)
3. \(15 = 3k\)
4. \(k = 5\)

Substitute \(k = 5\) back into the expression for the general term:

\(T_5 = \binom{9}{5} \left(\frac{4}{5}\right)^{4} \left(\frac{5}{2}\right)^5 x^{-6}\).

Calculate the coefficient:

• \(\binom{9}{5} = 126\)
• \(\left(\frac{4}{5}\right)^4 = \frac{256}{625}\)
• \(\left(\frac{5}{2}\right)^5 = \frac{3125}{32}\)

Combine these to find the coefficient:

\(126 \times \frac{256}{625} \times \frac{3125}{32}\)

1. Cancel \(\frac{3125}{625} = 5\), so the expression becomes \(126 \times 256 \times 5 / 32\)
2. Calculate \(126 \times 256 = 32256\)
3. Then \(32256 \times 5 = 161280\)
4. Finally, \(161280 / 32 = 5040\)

The coefficient of \(x^{-6}\) is \(5040\), which lies within the expected range (5040,5040).