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Lol xD

I tried it again to study for my math test tomorrow...I only got as far as [blank]/((n+1)(n+2)). Is there a way to write a nth term for something that pluses by another integer each time, as in 1, 3, 6, 10, 15, 21, etc?

Yes, there is. It usually involves an n². The particular sequence you described there is (n² + n)/2, or equivalently, n(n+1)/2, which I mentioned a few posts back and which I would be very surprised if you hadn't covered in class.

I guess then #38 would be (n(n+1))/(2(n+1)(n+2))

I as far as I recall, class hasn't covered anything divided by 2.
..I'm not entirely certain what has been covered in class except what sequences/sigmas/partial sums mean and that and looking for ratios and addition of similar numbers is important.

EDIT: How does (n² + n)/2 work?
Add a number the the same number multiplied by itself makes it always even..
n(n+1) means that with every increase in n, the "1" part increases by n too.
And then dividing by 2 makes it increase by 1 instead of 2.
So I guess I can instead of dividing it by 2, replace that with anything 2x = number to increase each time by that number? Only the first number will always get higher or lower. If I wanted it to stay at 1, then I'd have to subtract or add something extra?

That's not really the right way to go about it. It's good that you're developing those insights, but it doesn't quite generalize -- your conclusions are too specific to be especially useful.

A better strategy is to look at the changes in the sequence.

If your sequence is 1 1 1 1 1 (or any constant number), then there are no n's.

Otherwise, find the differences between the steps. 1 2 3 4 5 -> 1 1 1 1, for example, or 1 3 6 10 15 -> 2 3 4 5. Repeat this process until you get to a constant number. Each time you go through this process, the order of the polynomial increases by one -- that is, 1 2 3 4 5 has an n¹ term, 1 3 6 10 15 has an n² term, and 1 16 19 39 69 91 has an n³ term (if I didn't screw up my math). And it sometimes helps to start from 0 instead of 1.

Obviously it's possible for you to get stuck in a place where this technique doesn't work -- it only works for polynomials! If you've got a 2ⁿ in there, you'll instead need to consider the ratios between successive terms instead of the difference.

Thank-you very much for the n(n+1) thing! I had a math test today where I had to use (n-1)² + (n-1) (or I guess that's just n(n-1)?) to find some sequence. That is how to start from 0 right?

The period before the test I learned from someone in physics class that a_i = a₁r^{something to do with n} and I guess that works because when things are in ratios, it means that every next number is just that ratio multiplied again (hence the power)? And I'm not sure exactly why it's multiplied by the a₁ except that if the power were n-1 then it could be the first number..?

I'm not sure how I could've passed the math test without the knowledge of these things which I haven't learned in class though o_o'

It's multiplied by a₁ because k⁰ = 1.

That's one way to start from 0, yes. It's far from the ONLY way. In fact it's usually easier to start from 0 once you wrap your head around the idea. The n(n+1) works fine from 0 as it is because 0(0+1) = 0; we can clearly see that the sum of 0..n for n=0 is (drumroll please) 0.

Oh, so it's possible to just start n at 0 :o. Is there a way to indicate what to start n at? n₁=0?

Formally, you do it this way:


This means "the sum from i=1 to n of i is n(n+1)/2".

It would also be correct to write i=0 for this one.

Ohh I see, so S_n can be written using that funny looking sign to be more specific.

Thank-you very much!

The symbol is called "sigma" (it's an uppercase sigma, to be specific) and it represents "summation". An uppercase pi (Π) represents a product (that is, multiplication) used the same way.

The S_n notation is more general. You can use it to describe any sequence. It's also fairly common to use f(n) notation, especially for sequences that you want to find a continuous (in the sense of a continuous function) extension for.

The disadvantage of the more general notations is that they are also less useful. Summations and products have a number of very useful properties -- for example, Σkx = kΣx -- that do not hold for generalized sequences. (Note that this is true in general: the more generalized a form is, the fewer consistent properties that are available using it.)

Oh wow, I didn't know that uppercase pi (for multiplying do something to n numbers together(?)) existed.

I'm on spring break so I've not done any math xD (no homework)

Πx is just 1 x 2 x ... x n, just like the sum, except with multiplication.

So... This is relevant for what you were studying recently, Potironette:


A visual proof that 1/3+1/3^2+1/3^3+...=1/2

Ohh I remember my teacher splitting up pieces of paper trying to show that, though I only partly understood it. But why equals one half? Doesn't it never reach it?

There are techniques to actually find what the value would be if you went on to infinity. You'll study these in calculus. One that you could do right now would be to find the closed form representation of the sum (that was the S_n stuff) and plug infinity in for n.

The infinite sum converges to 1/2. The more terms you add, the closer you get.