Find the 7th term in the expansion of (4x/5+5/2x)^8

Question

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

Answer 1

To find: 7^th term in the expansion of \(\Big(\frac{4x}{5}+\frac{5}{2x}\Big)^8\)

Formula used: (i)

 ⁿC_r = \(\frac{n!}{(n-r)!(r)!}\)

(ii) T_r+1 = ⁿC_r a^n-r b^r 

For 7^th term, r + 1 = 7 

⇒ r = 6

In, \(\Big(\frac{4x}{5}+\frac{5}{2x}\Big)^8\)

7^th term = T₆₊₁