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

Explanation

Let ` T_(r +1)` be the greatest term in the expansion of
`(3 - 5)^(11) ` , we have
`(T_(r+1))/(T_(1)) = ((11- r +1)/(r))|- (5x)/(3)| `
` = ((12-r)/(r))|-(1)/(3)| = (12-r)/(3r) " " [ because x = 1 //5]`
` therefore (T_(r +1))/(T_(r)) ge 1 rArr (12 - r)/(3r) ge 1 rArr 12 ge 4r `
` therefore r le 3 rArr r = 2,3 `
So, the greatest terms are ` T_(2 + 1) " and " T_(3 +1)`
` therefore ` Greatest term (when r = 2 ) `= T_(2 + 1) = ""^(11)C_(2) (3)^(9) (-5x)^(2)`
` = (11*10)/(1*2) * 3^(9) * (1)^(2) = 55 xx 3^(9) " " [ because x = 1 //5]`
and greatest term when r = 3 ` T_(3+1) `
`= |""^(11)C_(3) (3)^(8) (-5x)^(3)| = |""^(11)C_(3) (3)^(8) (-1)^(3) | [ because x = 1//5]`
` = ""(11)C_(3) * 3^(8) = (11*10*9)/(1*2*3) . 3^(8) = 55xx3^(9)`
From above , we say that the values of both greatest terms are equal .
Aliter
Since , ` (3 - 6x)^(11) = 3^(11) (1 - (5x)/(3))^(11)`
Now , `m = ((11 +1)|-(5x)/(3)|)/(|- (5x)/(3)| + 1) = (12 xx|- (1)/(3)|)/(| - (1)/(3)| +1) " " [ because x = (1)/(5)]`
` = (4)/((1)/(3) + 1) = 3 `
Since , the greatest terms in the expansion are ` T_(3) " and " T_(4)` .