Question

Find the 7th term in the expansion of ${(\frac{4x}{5}+\frac{5}{2x})}^8$

Answer 1

General term in a binomial expansion

of $(a+b{)}^n$ is given by,

$Tr+1{=}^n{C}_r.a^r.b^{n-r}$

Here, $a=\frac{4x}{5},b=\frac{5}{2x},n=8$

For $7^{th}$ term, $r=6$

$\Rightarrow T_7{=}^8{C}_6\times \left(\frac{4x}{5}{)}^6\right(\frac{5}{2x}{)}^2$

$=\frac{8\times 7}{2}\times \frac{4^6}{5^5}\times x^6\times \frac{5^2}{2^2\times x^2}$

$=\frac{7\times 4^7}{5^4\times 4}\times x^4$

$=\frac{7}{5^4}\times 4^6\times x^4$

$T_7=\frac{28672}{625}{x}^4$