Question

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

Answer 1

Correct Answer: 5040

To solve the problem of finding the coefficient of \(x^{-6}\) in the expansion of \(\left(\frac{4x}{5} + \frac{5}{2x^2}\right)^9\), we use the multinomial expansion technique. In general, for expressions of the form \((a+b)^n\), the multinomial theorem gives us the term \(\binom{n}{k_1, k_2} a^{k_1} b^{k_2}\) where \(k_1+k_2=n\). Here, \(a=\frac{4x}{5}\) and \(b=\frac{5}{2x^2}\).

The term in the expansion that will potentially have \(x^{-6}\) can be represented as:

\[\binom{9}{k_1,k_2} \left(\frac{4x}{5}\right)^{k_1} \left(\frac{5}{2x^2}\right)^{k_2}\]

where \(k_1+k_2=9\). This expands to:

\[\binom{9}{k_1,k_2} \left(\frac{4}{5}\right)^{k_1} x^{k_1} \left(\frac{5}{2}\right)^{k_2} x^{-2k_2}\]

Simplifying the powers of \(x\), we need \(x^{k_1-2k_2} = x^{-6}\). Therefore:

\[k_1 - 2k_2 = -6\]

We have two equations:

\[k_1 + k_2 = 9\]

\[k_1 - 2k_2 = -6\]

Solving these equations simultaneously, subtract the second from the first:

\[3k_2 = 15 \Rightarrow k_2 = 5\]

Substitute \(k_2=5\) into the first equation:

\[k_1 + 5 = 9 \Rightarrow k_1 = 4\]

Thus, the relevant term in the expansion is when \(k_1=4\) and \(k_2=5\):

\[\binom{9}{4,5} \left(\frac{4}{5}\right)^4 \left(\frac{5}{2}\right)^5 x^{-6}\]

Calculating constants:

\[\binom{9}{4,5} = \frac{9!}{4!5!} = 126\]

\( \left(\frac{4}{5}\right)^4 = \frac{256}{625} \)

\( \left(\frac{5}{2}\right)^5 = \frac{3125}{32} \)

Thus, the coefficient is:

\[126 \times \frac{256}{625} \times \frac{3125}{32}\]

Calculating step by step:

\(\frac{256 \times 3125}{625 \times 32} = \frac{800000}{20000} = 40\)

Therefore, the coefficient is:

\[126 \times 40 = 5040\]

This satisfies the range \([5040,5040]\), confirming the calculated coefficient of \(x^{-6}\) is indeed \(5040\).