The algebraically second largest term in the expansion of `(3-2x)^(15)` at `x=(4)/(3)`. is (a) `5` (b)`7` (c)`9` (d)`11`

Options

A. `5`
B. `7`
C. `9`
D. `11`

Answer

B

Explanation

`(b)` `(3-2x)^(15)` at `x=(4)/(3)`
For greatest term, `r le (16|2x|)/(3+|2x|)=(16xx2xx(4)/(3))/(3+(8)/(3))=(128)/(17)`
`:.r=7`
`:.t_(r+1)=t_(8)=^(15)C_(7)3^(8)(-2x)^(7)=-^(15)C_(7)3^(8)((8)/(3))^(7)`
greates term
`:.t_(7)` and `t_(9)` are positive term
`t_(7)=^(13)C_(6)3^(9)(-2x)^(6)=^(15)C_(6)3^(9)(8//3)^(6)=^(15) C_(6)8^(6)*3^(3)`
and `t_(9)=^(15)C_(8)3^(7)(-2x)^(8)=^(15)C_(8)3^(7)(8//3)^(8)=^(15) C_(8)8^(8)*3^(-1)`
`:.t_(9)-t_(7)=(15!)/(8!7!)(8^(8))/(3)-(15!)/(8!6!)8^(6)*3^(3)=(15!)/(9!6!)(8^(6))/(3)[(64)/(7)-(3^(4))/(9)] gt 0`
`:.t_(7) lt t_(9)`
`t_(11)=^(15)C_(10)(8^(10))/(3^(5))`
`:.t_(7)-t_(11) gt 0`
`:.t_(7) gt t_(11)`
Thus `t_(9) gt t_(7) gt t_(11)`
Hence `t_(7)` is the second largest term.