Find numerically greatest term in the expansion of `(2 + 3 x)^9`, when x = 3/2.

Explanation

Let ` T_(r +1)` be the greatest term in the expansion of
` (2+ 3x)^(9)` , we have
`(T_(r+1))/(T_(r)) = ((9 - r + 1)/(r))|(3x)/(2)| = ((10-r)/(r)) |(3)/(2)xx(3)/(2)| = (90-9r)/(4r)" " [ because x = 3//2]`
`therefore (T_(r +1))/(T_(r)) ge 1`
` rArr (90 - 9r)/(4r) ge rArr 90 ge 13r `
` therefore r le (90)/(13) = 6 (12)/(13)`
or ` r le 6 (12)/(13)`
` therefore ` Maximum value of r is 6.
So , greatest term `= T_(6+1) = ""^(9)C_(4) (2)^(9-6) (3x)^(6)`
`= ""^(9_)C_(3) * 2^(3) (3xx(3)/(2))^(6)`
`= (9*8*7)/(1*2*3) (2^(3) * 3^(12))/(2^(6)) = (7xx 3^(12))/(2) `
Aliter Since , `(2 + 3x)^(9) = 2^(9) (1(3x)/(2))^(9)`
Now , `((9+1)|(3x)/(2)|)/(|(3x)/(2)| + 1)= (10xx(9)/(4))/((9)/(4) + 1)" " [ because x = 3 //2]`
` = T_([m]+1) = T_(6 + 1) "in" (2 + 3x)^(9)`
` = ""^(9)C_(6) (2)^(9-6) (3x)^(6) = ""^(9)C_(3) * 2^(3) ((3^(2))/(2))^(6)`
`(90)/(13) = 6 (12)/(13) ne ` Integer
` therefore ` The greatest term in the expansion is
` = (9*8*7)/(1*2*3). (3^(12))/(2^(3)) = (7xx3^(13))/(2)`