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Find the middle term in the expansion of `(ax-b/x^(2))^(12)` |
Answer» 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))) xx(924a^(6)b^(6))/x^(6).` Hence, the middle term in the given expansion is `(924a^(6)b^(6))/x^(6)`. |
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