Find the middle term in the expansion of `(ax-b/x^(2))^(12)`

Explanation

The general term in the given expansion is given by `T_(r+1)=(-1)^(r) xx ^12C_(r) xx (ax)^((12-r)) xx (b/x^(2))^(r)`
` rArr T _(r+1)= (-1)^(r) xx ^12C_(r) xxa^((12-r)) xx b^(r) xx x^((12-3r))" " `.
Since the given binomial contains an even power, so it has only one middle term.
Middle term `=(12/2+1)`th term =7th term.
Now, `T_(7)=T_((6+1))`
`(-1)^(6) xx ^(12)C_(6) xx a^((12-6)) xx b^(6)xx x^((12-3xx6))" "` [putting `r = 6(i)` ].
`.= ^(12)C_(6) xxa^(6)b^(6) xxx^(-6)`
` = ((12xx11xx10xx9xx8xx7)/(6xx 5xx4xx 3xx2xx1) xx (a^(6)b^(6))/(x^(6))) =(924a^(6)b^(6))/x^(6).`
Hence, the middle term in the given expansion is `(924a^(6)b^(6))/x^(6)`.