Recurring Dreams!

Algebra Level 5

y n + 1 = 10 ⌊ 4 y n − 3 ⌋ \large{{ y }_{ n+1 }=10\left\lfloor \sqrt { 4{ y }_{ n }-3 } \right\rfloor }

If there be a sequence defined by above rule for n ≥ 1 n \ge 1 . Find y 2015 y_{2015} if y 1 = 9000 y_{1}=9000


The answer is 390.

Department 8 by Department 8 , 27, Israel

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

Bob Kadylo
May 21, 2017

The first 9 terms of this sequence are: 9000,1890,860,580,480,430,410,400,390. From that point to Infinity, the sequence terms are all 390 because 10 ⌊ 4 ( 390 ) − 3 ⌋ = 390 10\lfloor\sqrt{4(390)-3}\rfloor=390 again.

0 Helpful 1 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...