If `x=1//3,` find the greatest tem in the expansion of `(1+4x)^8dot`

Answer

`T_(6) = 56(4/3)^(5)`

Explanation

We have
Now, `(T_(r+1))/(T_(r)) = (n-r+1)/(r ) (4x) = (9-r)/(r ).(4)/(3)` (Putting `n = 8, x=1//3`)
`:. (T_(r+1))/(T_(r)) ge 1`
`rArr 36-4r ge 3r`
`rArr 36 ge 7r`
`rArr r le 51/7`
Hence `r = 5`, and therefore, `T_(5+1)` i.e., `6^(th)` term is greatest.
`rArr T_(6) = .^(8)C_(5)(4x)^(5) = (8!)/(5! xx 3!) .(4/3)^(5)`
`= (8xx7xx6)/(6) (4/3)^(5)`
`= 56(4/3)^(5)`