Find the numerically grates term in the expansion of `3-5x^(15)w h e nx=1//5.`

Options

A. `""^(15)C_(3) xx3^(10)`
B. `""^(15)C_(3) xx3^(11)`
C. `""^(15)C_(12) xx3^(12)`
D. `""^(15)C_(11) xx3^(12)`

Answer

c

Explanation

Let `T_(r +1) and T_(r)` denote the `(r +1)^(th)` term
respectively. Then
`T_(r-1) = ""^(15)C_(r) 3^(15-r) (-5x)^(r) and , T_(r) = ""^(15)C_(r-1)3^(15 - r +1) (-5x)^(r-1)`
`rArr (T_(r+1))/(T_(r)) = (15-r+1)/(r)((-5x)/(3))`
`rArr (T_(r+1))/(T_(r)) = (16-r)/(r)xx(-(5)/(3)xx(1)/(5))`, when `x = (1)/(5)`
`rArr (T_(r+1))/(T_(r)) = (16-r)/(r)xx(1)/(3) `, numberically [ Neglecting minus sing]
Now,
`(T_(r) +1)/(T_(r)) xx(1)/(3) gt 1 rArr 16 gt 4r rArr r lt 4`
Since 4 is an integer. Therefore , 4th and 5th terms are
numberically greatest terms.
Now ,
`T_(4) = T_(3+1) = ""^(15)C_(3) = ""^(15)C_(3) 3^(15-3) (-5x)^(3)`
`rArr T_(4) = ""^(15)C_(3) xx3^(12) (-5xx(1)/(5))^(3), when. x = (1)/(5)`
`rArr T_(4) = ""^(15)C_(3) xx3^(12)` (numberically)