Determine N, given that:

2014
Σ Arctan (k²+k+1)^-1 = Arctan (N)
k=0

*** All angles are in radians.

To see what's going on, replace the x=2014 with smaller numbers.
x=0
tan(arctan(1))=1

x=1
tan(arctan(1)+arctan(1/3))=2 [this was a recent problem solved via origami]

x=2
tan(arctan(1)+arctan(1/7))=3

So there's a pattern here.  The answer would seem to be N=2015

The addition of a single term seems to make the answer go up by 1.  Lets see what happens when we work backwards -- try to make the answer go up by one to see what the term must be.
My calculator says tan(arctan(4)-arctan(3))=1/13
which is promising.  That's 3²+3+1
Lets try
tan(arctan(x+1)-arctan(x)) = [(x+1)-(x)]/[1+(x+1)(x)] = (x²+x+1)^-1
Bingo!
So this is the term that makes the answer increase by 1.
So I've proven N=x+1
So when x=2014, N=2015.

Posted by Jer on 2014-07-01 11:17:51