2015 once again

Algebra Level 2

Let x x be a non-real number such that x 2 + x = − 1 x^{2}+x=-1 . Find the value of ∑ n = 0 2015 ( x 3 n ) \displaystyle \sum_{n=0}^{2015} (x^{3n})


The answer is 2016.

Hjalmar Orellana Soto by Hjalmar Orellana Soto , 24, Chile

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.

2 solutions

x 2 + x + 1 = 0 x^{2}+x+1=0 = > x 3 = 1 =>x^{3}=1 = > ∑ n = 0 2015 ( x 3 n ) = ∑ n = 0 2015 1 = 2016 =>\displaystyle \sum_{n=0}^{2015} (x^{3n})=\displaystyle \sum_{n=0}^{2015} 1=2016

1 Helpful 0 Interesting 0 Brilliant 0 Confused 
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
 
x**2 + x + 1 = 0

multiply both sides by (x-1)

(x-1)(x**2 + x + 1) = 0
x**3 - 1**3 = 0
x**3 = 1

from  n=1 to n=2015:
1*2015 = 2015
( 1^n = 1 )

for n=0:
1^0 = 1

2015+1 = 2016

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...