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InterviewSolution
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Find the numerically Greatest Term In the expansion of `(3-5x)^15` when x=1/5A. `""^(15)C_(3) xx3^(10)`B. `""^(15)C_(3) xx3^(11)`C. `""^(15)C_(12) xx3^(12)`D. `""^(15)C_(11) xx3^(12)` |
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Answer» Correct Answer - c Let `T_(r +1) and T_(r)` denote the `(r +1)^(th)` term respectively. Then `T_(r-1) = ""^(15)C_(r) 3^(15-r) (-5x)^(r) and , T_(r) = ""^(15)C_(r-1)3^(15 - r +1) (-5x)^(r-1)` `rArr (T_(r+1))/(T_(r)) = (15-r+1)/(r)((-5x)/(3))` `rArr (T_(r+1))/(T_(r)) = (16-r)/(r)xx(-(5)/(3)xx(1)/(5))`, when `x = (1)/(5)` `rArr (T_(r+1))/(T_(r)) = (16-r)/(r)xx(1)/(3) `, numberically [ Neglecting minus sing] Now, `(T_(r) +1)/(T_(r)) xx(1)/(3) gt 1 rArr 16 gt 4r rArr r lt 4` Since 4 is an integer. Therefore , 4th and 5th terms are numberically greatest terms. Now , `T_(4) = T_(3+1) = ""^(15)C_(3) = ""^(15)C_(3) 3^(15-3) (-5x)^(3)` `rArr T_(4) = ""^(15)C_(3) xx3^(12) (-5xx(1)/(5))^(3), when. x = (1)/(5)` `rArr T_(4) = ""^(15)C_(3) xx3^(12)` (numberically) |
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