please help with this challenging math problem?

Question

the problem is: Find real constants A,B, C such that ((1/x^3)-x)=(A/x)+(B/x-1)+(C/x+1).
So A = -1, B = 1/2, C = 1/2
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now im supposed to use that problem to solve this problem. It contains symbols i cant type so i loaded it on photobucket
http://i230.photobucket.com/albums/ee31/meow_meoww/EEEE.jpg
plz show work. I really need help on this problem. i dont get it at all.thanks.

Answer 1

I take it you mean 1/(x^3 - x) = A/x + B/(x-1) + C/(x+1) ?

It looks like you already solved the first problem. Now use this to rewrite the sum:

∑ 1/(k^3 - k) =
∑ [ -1/k + 1 /2(k-1) + 1 /2(k+1) ] =
∑ -1/k + (1/2)∑1/(k-1) + (1/2)∑ 1/(k+1) =
-∑ 1/k + (1/2)∑1/(k-1) + (1/2)∑ 1/(k+1)

Now each of these sums go from 2 to 2007, but we can replace the k-1 with k and take the sum from 1 to 2006, and likewise replace the k+1 with k and take that sum from 3 to 2008:

-∑[2to2007] 1/k + (1/2)∑[1to2006]1/k + (1/2)∑[3to2008]1/k

Right now just look at the last two terms:
(1/2) [ ∑[1to2006]1/k + ∑[3to2008]1/k ]
(1/2) [ ∑[1to2006]1/k + ∑[3to2006]1/k + 1/2007 + 1/2008 ]
(1/2) [ ∑[1to2006]1/k + ∑[1to2006]1/k + 1/2007 + 1/2008 - 1/1 - 1/2]
(1/2) [ 2∑[1to2006]1/k + 1/2007 + 1/2008 - 1/1 - 1/2]
∑[1to2006]1/k + 2/2007 + 2/2008 - 3

Remember the first term was -∑[2to2007]1/k. Rewrite this as:
-∑[1to2007] 1/k + 1/1
-∑[1to2006] 1/k + 1/1 - 1/2007

Add this to the other sum we got, and the big Sigma terms cancel! You get
2/2007 + 2/2008 - 3 + 1 - 1/2007
which you can simplify from there

Answer 2

The only weird symbol is the big Greek "SIGMA", which is the symbol for a summation, and in this case from k=2 to k=2007. So the deal is that you have the same type term inside the "SIGMA". So you are going to break it up into 3 fractions like the ones you solved for, and each will be in a SIGMA. I don't think you need to solve each summation.

Answer 3

ok shall we see... instruct that for each non-adverse integer n, the selection 7^7^n +one million is the made of a minimal of 2n+3 (no longer unavoidably distinctive) top factors. This became from the 2007 USAMO. on the competition, I only wrote for n=0, the project is trivial. something of the evidence by employing induction is left as an workout for the grader. LOL