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Summing square inverses (Posted on 2011-12-25) Difficulty: 3 of 5
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.

See The Solution Submitted by K Sengupta    
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Solution Pen and paper solution Comment 2 of 2 |
First lets fix the format of the individual terms:

(1+n-2+(n+1)-2)1/2
=((n(n+1) + (n+1) + n)/(n(n+1)))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
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