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Help---question from Princeton GMAT Math Bible

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by EmilKhalikov » Fri Dec 19, 2008 9:17 am
1 + 2 + 3 + ... + n = n(n+1)/2
1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6
1^3 + 2^3 + ... + n^3 = ( n^2(n+1)^2 )/4

I cannot believe they would ask such question. If you do not remember the formulas, you are pretty much screwed because it is nearly impossible to figure out those formulas.
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by amyyoung999 » Fri Dec 19, 2008 12:07 pm
Thank you very much!
I encountered this question when I took the test last month,and my friend
encountered the same question this week.
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by EmilKhalikov » Fri Dec 19, 2008 12:17 pm
Here is more:
1 + 3 + 5 + ... + (2n - 1) =n^2
2 + 4 + 6 + ... + 2n = n*(n+1)
1^2 + 3^2 + ... + (2n-1)^2 = n(4n^2 - 1)/3
1^3 + 3^3 + ... + (2n-1)^3 = n^2(2n^2 - 1)

Combining these formulas with the ones I posted earlier should help calculating various integer series
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by amyyoung999 » Fri Dec 19, 2008 3:16 pm
Wow!You are an elite on Math.
You must hold a PH.D from department of Mathematics.
Thanx a lot!
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by EmilKhalikov » Fri Dec 19, 2008 8:27 pm
Lol no, I am Applied Math major and I am pretty good at it, so I figure I help other people. It's a good practice for me too.
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by amitabhprasad » Sat Dec 20, 2008 1:04 am
@ amyyoung999
do you mean you go this test in GMATPrep questions ?
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by amyyoung999 » Sat Dec 20, 2008 6:32 am
This question is from Princeton GMAT Math Bible.
I relize that the questions on the Princeton GMAT Math Bible are much more difficult than the questions on GMATPREP.
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