Question:
Find the 7 th term in the expansion of $\left(\frac{4 x}{5}+\frac{5}{2 x}\right)^{8}$.
Solution:
We need to find the 7th term in the given expression.
$\because T_{7}=T_{6+1}$
$\therefore T_{7}=T_{6+1}$
$={ }^{8} C_{6}\left(\frac{4 x}{5}\right)^{8-6}\left(\frac{5}{2 x}\right)^{6}$
$=\frac{8 \times 7 \times 4 \times 4 \times 125 \times 125}{2 \times 1 \times 25 \times 64} x^{2}\left(\frac{1}{x^{6}}\right)$
$=\frac{4375}{x^{4}}$