Hey, I'm new to this forum and I think the website is awesome. I was wondering about two particular questions. First, on the test I have noticed several different series of 1^3-2^3+3^3-4^3+5^3 and so on. I was wondering the formula to such problems. Also, I was wondering what practice tests you advise to take. Currently I'm working the UIL tests from the last three years, the 990 and 989 series, and 2003-2004 from Dr. Numsen. What advice do you have for preparing for regional? Thanks and I love the site.
Thank you. It is always good to know the countless hours I put into the website are being put to good use.
I know I have been out of number sense too long. It has been 8 years and most of the questions I have gotten in this forum are for tricks I have never seen. Your question is no exception. I have never seen this trick.
I have been trying to come up with a way of doing this quickly, but to no avail. So if anyone has a particular strategy you use, please share with us.
I did come up with 2 formulas that do work. It is almost just as easy to do the adding/subtracting individually though. But I will let you decide.
If you are looking for the nth sum, then we have 2 situations. If n is even or if n is odd. So let's look at n being even. (Please don't ask how I came up with this formula, I am not even totally sure myself, but it does work.)
If n is even use:
-[n^2(2n+3)]/4
Ex [1] 1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 = ___
In this example n = 6.
Using the formula we get -36(15)/4 = -9*15 = -135.
The answer is -135.
If n is odd use (this is a little harder):
(2n^3 + 3n^2 - 1)/4 or if it is easier [n^2(2n+3) - 1]/4
Ex [2] 1^3 - 2^3 + 3^3 - 4^3 + 5^3 = ____
In this example n = 5.
Using the formula we get (250 + 75 - 1)/4 = 324/4 = 81 [or (25*13 - 1)/4 = 324/4].
The answer is 81.
I am sorry that I can give you nothing simpler at this time. Like I said, if you don't want to memorize 2 new formulas, you can work it out individually which actually doesn't take that long to do. Or you can use this method. It is up to you.
If anyone has any other ways to do this problem, I would greatly appreciate your method.
As for your second question the answer is "I don't know!" I have been out of the number sense world for 8 years and I am not even sure what is available. I have seen other forums where people talk a lot about Dr. Numsen's tests. The truth is, any test will be beneficial.
The way I studied was I did LOTS of tests, maximum 5 a day. (You don't want to burn yourself out.) But I did not stop there. I was a bit of a nerd and created my own problems to help me. I knew which problems slowed me down or which problems I had most difficulty with, so I would create a test with lots of examples of these types of problems. By doing this, I was able to gain lots of speed on slower problems. So you can try this too. Just find what you are not great at and practice it.
Everyone learns in different ways. For me, taking tests was only a measure of how much I learned, not how I learned (if that makes sense). Other people, that is all they use and they do fine.
I hoped I helped a little. Come back if you find other questions. I enjoy working on them (especially new problems I have never seen).
Well, I did some more work on your problem and came up with a much better, much easier, much nicer way of doing it. And you have one basic formula (with only a little alteration):
In this case, the problem is switched where the odds are negative which only negates the negative from our equation.
a = 3 since 6/2 = 3.
3^2 * (4(3) + 3) = 9 x 15 = 135.
The answer is 135.
Well, now I feel a lot better. I knew there had to be an easier way of doing these problems. I hoped this answered your question and ignore my original post.
Awesome. That's a pretty convienient formula for solving those series. I could never seem to figure out a way to do them. Thanks for the awesome new strategy.
