Get ready Part 31

Calculus Level 3

Consider the power series expansion 1 1 − 2 x − x 2 = ∑ n = 0 ∞ a n x n \dfrac{1}{1-2x-x^2}=\displaystyle \sum_{n=0}^{\infty} a_nx^n . Is it true that for each integer n ≥ 0 n \ge 0 , there is an integer m m such that a n 2 + a n + 1 2 = a m a^2_n+a^2_{n+1}=a_m ?

No Yes

Gia Hoàng Phạm by Gia Hoàng Phạm , 19, Vietnam

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1 solution

Mark Hennings
May 27, 2019

Expanding 1 1 − 2 x − x 2 \frac{1}{1-2x-x^2} we deduce that a n = 1 2 2 [ ( 1 + 2 ) n + 1 − ( 1 − 2 ) n + 1 ] a_n \; = \; \frac{1}{2\sqrt{2}}\left[\big(1 + \sqrt{2}\big)^{n+1} - \big(1 - \sqrt{2}\big)^{n+1}\right] from which it is easy to show that a n 2 + a n + 1 2 = a 2 n + 2 a_n^2 + a_{n+1}^2 = a_{2n+2} for all n ≥ 0 n\ge 0 .

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