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

Answer

`T_(7) = (7 xx3^(13))/2`

Explanation

We have , `(2+3x)^(9) = 2^(9) (1+(3x)/2)^(9).`
`T_(r+1) = 2^(9) xx .^(9) C_(r) ((3x)/2)^(r) and T_(r) = 2^(9) xx .^(9)C_(r-1)((3x)/2)^(r-1).`
`:. (T_(r+1))/(T_(r)) = 2^(9) xx .^(9)C_(r) xx 3^(r)/2^(r) xx x^(r) xx (2^(r-1))/(3^(r-1)) xx 1/(x^(r-1)) xx1/(2^(9) xx .^(9) C _(r-1))`
`=3/2 xx x xx (.^(9)C_(r))/(.^(9)C_(r-1)) = (3/2x) xx ((9!))/((r!)*(9-r)!) xx((r-1)!)/((9!)) xx ( 10 - r)!.`
`=(3/2 xx3/2) xx ((10-r))/r= (9(10-r))/(4r)" "[:' x= 3/2]` ,
Let the greatest term be `T_(r+1)`.
Then , `T_(r+1) ge T_(r) rArr (T_(r+1))/(T_(r)) ge 1 rArr (9(10-r))/(4r) ge 1.`
`rArr 4r le 90 - 9r rArr 13r le 90 rArr r le 6 (12)/13.`
Thus, the maximum value of r is 6 for which `T_(r+1)` is greatest.
` :. " greatest term " = T_(6+1) = 2^(9) xx .^(9)C_(6) xx (3^(6))/(2^(6)) xx x^(6), " where " x = 3/2.`
` :. " numerical value of greatest term at " x= 3/2 = 2^(9) xx .^(9) C_(3) xx (3^(6))/(2^(6)) xx (3^(6))/(2^(6)) `
`(9xx 8 xx 7 )/(3 xx 2 xx 1) xx (3^(12))/(2_(3))= ( (7 xx 3^(13))/2).`