Let α and β be the roots of x 2 - x - 1 = 0 , with α > β . For all positive integers n , define a n = α n - β n α - β ,   n ≥ 1 b 1 = 1 and b n = a n - 1 + a n + 1 ,   n ≥ 2 . Then which of the following options is/are correct?
D. b n = α n + β n for all n ≥ 1
Given, α and β are roots of x 2 - x - 1 =   0 a k + 2 - a k = α k + 2 - β k + 2 - α k - β k α - β = α k + 2 - α k - β k + 2 - β k α - β = α k α 2 - 1 - β k β 2 - 1 α - β = a k + 1 ⇒ a k + 2 - a k = a k + 1   ⇒ a k + 2 - a k + 1 = a k ∑ a k k = 1 n = a k + 2 - a 2   = a k + 2 - α 2 - β 2 α - β = a k + 2 - α + β = a k + 2 - 1 ⇒ a 1 + a 2 + a 3 + . . . . . + a n = a n + 2 - 1 Option ( B ) ∑ n = 1 ∞ a n 10 n = ∑ n = 1 ∞ α n - β n ( α - β ) 10 n = 1 α - β ∑ n = 1 ∞ α 10 n - β 10 n = 1 α - β α 10 1 - α 10 - β 10 1 - β 10 = 1 α - β α 10 - α - β 10 - β = 1 α - β 10 α - 10 β 10 - α 10 - β = 1 α - β 10 α - β 100 - 10 α + β + α β = 10 100 - 10 - 1 = 10 89 Option ( C ) : ∑ n = 1 ∞ b n 10 n = ∑ n = 1 ∞ α n + β n 10 n = ∑ n = 1 ∞ α 10 n + β 10 n = α 10 1 - α 10 + β 10 1 - β 10 apply sum of infinite G.P. with a = α 10   a n d   r = α 10 = 10 α + β - 2 α β 10 - α 10 - β = 10 ( 1 ) - 2 ( - 1 ) 100 + α β - 10 ( α + β ) = 12 100 - 1 - 10 = 12 89 Option ( D ) : As given b n = a n - 1 + a n + 1 b n = α n - 1 - β n - 1 α - β + α n + 1 - β n + 1 α - β ∵ α - β = 1 b n = α n - 1 - β n - 1 + α n + 1 - β n + 1 Now, as α β =   1 ∴ α n β = - α n - 1 ∴    α β n = - β n - 1 b n = α n α - β + β n α - β b n = α n + β n
Related: Mathematics — Quadratic Equation · All PYQ Banks