For real numbers (a/b/c) define s_n =a^n +b^n +c^n Suppose s_1 =2 s_2 =6 and s_(3 ) =14.Prove that ∣s_n ^2 −s_(n−1) s


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Question Number 123595 by Snail last updated on 26/Nov/20

$F o r r e a l n u m b e r s (a / b / c) d e f i n e s_{n} = a^{n} + b^{n} + c^{n}$ $S u p p o s e s_{1} = 2 s_{2} = 6 a n d s_{3} = 14 . P r o v e t h a t$ $∣ s_{n}^{2} - s_{n - 1} s_{n + 1} ∣= 8 = h o l d s \forall n > 1$
Commented bySnail last updated on 26/Nov/20

$I h a d t r i e d u s i n g M a t h e m a t i c a l I n d u c t i o n b u t u n a b l d t o s o l v e$
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