InterviewSolution
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InterviewSolution
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Let `p, q` be integers and let `alpha,beta` be the roots of the equation `x^2-2x+3=0` where `alpha != beta` For `n= 0, 1, 2,.......,` Let `alpha_n=palpha^n+qbeta^n` value `alpha_9=`A. `a_(n) + 1 = a_(n) + a_(n) -1`B. `a_(n) + 2 = a_(n) +1 + a_(n) -1`C. `a_(n)+1 = a_(n) + 1`D. `a_(n)+1 = a_(n) -1 + 1` |
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Answer» Correct Answer - A It is given that `alpha, beta` are roots of `x^(2) - x - 1 = 0`. `therefore" "alpha + beta = 1 and alpha beta = - 1`. We have, `a_(n) = p alpha^(n) + q beta^(n)` `therefore" "a_(n+1) = p alpha^(n+1) + q beta^(n+1)` `rArr" "a_(n+1) = (p alpha^(n) + q beta^(n)) (alpha+beta) - alpha beta (p alpha^(n-1) + q beta^(n-1))` `rArr" " a_(n+1) = a_(n) + a_(n-1)" "[therefore alpha+beta = 1, alpha beta = - 1]` Hence, option (a) is correct. |
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