Determine the value of :

(1+1^-2+2^-2)^1/2 + (1+2^-2+3^-2)^1/2 +......+ (1+2010^-2+2011^-2)^1/2

*** For an extra challenge, solve this puzzle using only pen and paper.

First lets fix the format of the individual terms:

(1+n^-2+(n+1)^-2)^1/2
=((n²(n+1)² + (n+1)² + n²)/(n²(n+1)²))^1/2
=((n^4 + 2n^3 + 3n^2 + 2n + 1)/(n^2(n+1)^2))^(1/2)
=(n^2 + n + 1)/(n^2 + n)
=1 + 1/(n^2 + n)
So each term is 1 plus a small fraction.  We can use partial fraction decomposition on this fraction so each term is now
=1 + 1/n - 1/(n+1)

From this we see that in successive terms the 1/(n+1) term will cancel the 1/n term of the previous so that if we add up a bunch of successive terms we get
[number of terms] + 1/(first n) - 1/(last n + 1)

The problem requests the sum of terms 1 to 2010 so the sum is
2010 + 1/1 - 1/2011
= 2011 - 1/2011
= 2010 + 2010/2011

Posted by Jer on 2011-12-25 17:23:09