Tessellate S.T.E.M.S (2019) - Mathematics - Category B - Set 2 - Objective Problem 2

Algebra Level 2

Given that x 1 = 211 x_1 = 211 ; x 2 = 375 x_2 = 375 ; x 3 = 420 x_3 = 420 ; x 4 = 523 x_4 = 523 ; and x n = x n − 1 − x n − 2 + x n − 3 − x n − 4 x_n = x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4} for n ≥ 5 n \geq 5 . Find x 531 + x 753 + x 975 x_{531} + x_{753}+x_{975}

This problem is a part of Tessellate S.T.E.M.S. (2019)

898 747 998 267

Tessellate S.T.E.M.S. Mathematics by Tessellate S.T.E.M.S. Ma… , 26

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Patrick Corn
Jan 18, 2019

Note that x n + 1 = x n − x n − 1 + x n − 2 − x n − 3 x_{n+1} = x_n - x_{n-1} + x_{n-2} - x_{n-3} and rearranging the original recurrence gives x n − x n − 1 + x n − 2 + x n − 3 = − x n − 4 . x_n-x_{n-1}+x_{n-2}+x_{n-3} = -x_{n-4}. So x n + 1 = − x n − 4 . x_{n+1} = -x_{n-4}. That is, x n + 5 = − x n x_{n+5} = -x_n for any n ≥ 1. n \ge 1. This implies that x n + 10 = x n . x_{n+10} = x_n. So x 531 = x 1 = 211 , x_{531} = x_1 = 211, x 753 = x 3 = 420 , x_{753} = x_3 = 420, x 975 = x 5 = 523 − 420 + 375 − 211 = 267 , x_{975} = x_5 = 523-420+375-211 = 267, and they add up to 898 . \fbox{898}.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...