InterviewSolution
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InterviewSolution
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Let m be the smallest positive integer such that the coefficient of `x^2` in the expansion of `(1+x)^2 + (1 +x)^3 + (1 + x)^4 +........+ (1+x)^49 + (1 + mx)^50` is `(3n + 1) .^51C_3` for some positive integer n. Then the value of n isA. 16B. 5C. 21D. 11 |
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Answer» Correct Answer - b We find that `(1 + x)^(2) + (1 + x)^(3) +… (1 + x)^(49) + (1 + mx)^(50)` ` = (1 + x)^(2) { ((1 + x)^(48) -1)/((1 + x) -1)} +(1 + mx)^(50)` ` (1)/(x) {( 1 + x)^(50) - (1 + x)^(2)} + (1 +mx)^(50)` Coeffi. Of `x^(2)` in `(1 + x)^(2) + (1 + x)^(3) + ...(1 + x)^(49) + (1 +mx)^(50)` = Coeffi.of `x^(2)` in `[(1)/(x) {(1 +x)^(50) - (1 + x)^(2)} +(1 +mx)^(50)]` ` Coeffi. of `x^(3)` in `{ (1 + x)^(50) - (1 + x)(2)} + ` Coefficient of `x^(2)` in `(1 +mx)^(50)` `""^(50)C_(3) + ""^(50)C_(2) m^(2)` It is given that this coefficient is `(3n +1) ""^(51)C_(3)` . `therefore ""^(50)C_(3) + ""^(50)C_(2) m^(2) = (3n +1)""^(51)C_(3)` `rArr (50xx49xx48)/(3xx2xx1)+(50xx49)/(2xx1)m^(2) = (3n-1)(51xx50xx49)/(3xx2xx1)` `rArr 16 + m^(2) = 17 (3n +1)` `rArr m^(2) = 51n+1` The least value of n for which `51 N+1` is a perfect square is 5 and for this value of n the value of m is 16. Hence,m=16 and n=5. |
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