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Find the middle term inn (i) ((2y^(2))/(3)+(3)/(2y^(2)))^(9),y ne0 (ii) (4x^(2)+9y^(2)+12xy)^(n) |
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Answer» Solution :(i) Number of terms in the expansion is (9+1)=10 (EVEN) `therefore` There are two middle terms given by `T_((9+1)/(2)) annd T_((9+3)/(2))` i.e., `T_(5) and T_(6)` We know that `T_(r+1)=.^(n)C_(r)x^(n-r)a^(r)` `thereforeT_(5)=T_(4+1)=.^(9)C_(4)((2y^(2))/(3))^(5)((3)/(2y^(2)))^(4)` `=(9xx8xx7xx6)/(4xx3xx2xx1)xx(2^(5))/(3^(5))y^(10).(3^(4))/(2^(4)y^(8))` `=(9xx8xx7xx6)/(4xx3xx2xx1)xx(2)/(3)xxy^(2)` `=84y^(2)`. and `T_(6)=T_(5+1)=.^(9)C_(5)((2y^(2))/(3))^(4)((3)/(2y^(2)))^(5)` `=(9xx8xx7xx6)/(4xx3xx2xx1)xx(2^(4))/(3^(4))xx(3^(5))/(2^(5))xx(1)/(y^(2))` `=(9xx8xx7xx6)/(4xx3xx2xx1)xx(3)/(2)xx(1)/(y^(2))` `=(189)/(y^(2))` (ii) `(4x^(2)+9y^(2)+12xy)^(n)` Since `4x^(2)+9y^(2)+12xy=(2x+3y)^(2)`, therefore `(4x^(2)+9y^(2)+12xy)^(n)=(2x+3y)^(2n)` Hence, there is only one middle TER namely `((2n)/(2)+1)^(th)` term i.e., `(n+1)^(th)` term. `therefore` The required middle term `=T_(n+1)=.^(2n)C_(n)(2x)^(2n-n)(3y)^(n)` . . . `=.^(2n)C_(n)(2x)^(n)(3y)^(n)` `=(lfloor(2n))/(lfloor(n)lfloor(n)).2^(n).3^(n)x^(n).y^(n)` hence, the middle term int he expansion of `(4x^(2)+9^(y)+12xy)^(n)` is `(lfloor(2n))/(floor(n)floor(n)).2^(n).3^(n)x^(n).y^(n)` |
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