InterviewSolution
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InterviewSolution
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In the expansion of `(x^3-1/(x^2))^n ,n in N`, if the sum of the coefficients of `x^5a n dx^(10)`, then `n`isa. 25 b. 20 c. 15 d. none of these |
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Answer» Correct Answer - A `(x^(3)-(1)/(x^(2)))^(n)` General term, `T_(r+1)=.^(n)C_(r)(x^(3))^(n-r).(-(1)/(x^(2)))^(r)` `=.^(n)C_(r).x^((3n-3r)).(-1)^(r).x^(-2x)` `=.^(n)C_(r).(-1)^(r).x^((3n-5r))" "...(i)` For the coefficient `x^(5)` Put 3n - 5r = 5 5r = 3n - 5 `therefore r = (3n)/(5)-1` `therefore "Coefficient of" x^(5)=.^(n)C_(((3n)/(5)-1))(-1)^(((3n)/(5)-1))` For the cefficient of `x^(10)` Put 3n - 5r = 10 5r = 3n - 10 `therefore r = (3n)/(5)-2` `therefore "Coefficient of" x^(10)=.^(n)C_(((3n)/(5)-2))(-1)^(((3n)/(5)-2))` The sum of the coefficient of `x^(5) and x^(10)=0` `rArr .^(n)C_(((3n)/(5)-1))(-1)^(((3n)/(5)-1))+.^(n)C_(((3n)/(5)-2))(-1)^(((3n)/(5)-2))=0` `rArr (-1)^((3n)/(5))[.^(n)C_(((3n)/(5)-1)).(-1)^(-1)+.^(n)C_(((3n)/(5)-2)).(-1)^((-2))]=0` `rArr -.^(n)C_(((3n)/(5)-1))+.^(n)C_(((3n)/(5)-2))=0" "...(ii)` n = 15 Total term in the expansion of `(x^(3)-(1)/(x^(2)))^(15)` is 16. `therefore` middle term = `8^(th)` term and `9^(th)` term `T_(8)=T_((7+1))=.^(15)C_(7).(-1)^(7).x^((3 xx 15 - 5 xx 7))` `=-.^(15)C_(7).x^(10)" "("from eq. (i)")` `T_(9)=T_((8+1))=.^(15)C_(8).(-1)^(8).x^((3 xx 15 - 5 xx 8))` `=-.^(15)C_(8).x^(5)" "("from eq. (ii)")` The sum of the coefficients of the two middle terms `=-.^(15)C_(7)+.^(15)C_(8)=-.^(15)C_(7)+.^(15)C_(7)." "[because .^(n)C_(r)=.^(n)C_(n-r)]` = 0 |
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