InterviewSolution
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InterviewSolution
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If the coefficient of 2nd, 3rd and 4thterms in the expansion of `(1+x)^(2n)`are in A.P. , show that `2n^2-9n+7=0.` |
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Answer» Given expansion is `(1 + x)^(2n)` Now, coefficient of 2nd term `= .^(2n)C_(1)` Coefficient of 3rd term `= .^(2n)C_(2)` Coefficient of 4th term `= .^(2n)C_(3)` Given that, `.^(2n)C_(1), .^(2n)C_(2) " and " .^(2n)C_(3)` are in AP. Then, `2. .^(2n)C_(2) = .^(2n)C_(1) + .^(2n)C_(3)` `rArr 2 [(2n (2n - 1) (2n - 2)!)/(2 xx 1 xx (2n - 2)!)] = (2n (2n -1)!)/((2n - 1)!) + (2n (2n -1) (2n - 2) (2n - 3)!)/(3! (2n - 3)!)` `rArr n(2n - 1) = n + (n(2n - 1) (2n -2))/(6)` `rArr n(12n - 6) = n (6 + 4n^(2) - 4n - 2n + 2)` `rArr 12 n - 6 = (4n^(2) - 6 n + 8)` `rArr 6(2n - 1) = 2 (2n^(2) - 3n + 4)` `rArr 3(2n - 1) = 2n^(2) - 3n + 4` `rArr 2n^(2) - 3n + 4 - 6n + 3 = 0` `rArr 2n^(2) - 9n + 7 = 0` |
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