Memorization, a contemptuous, futile, and insignificant method of learning math is, I think, not the solution to my problem. Well for the arithmetical progression, I have the formula: Let: l= last term a= first term d= common difference l= a+[d(n-1)] And for the geometrical progression; Let: l= last term a= first term r= common ratio l= a(r^n-1) and also, for the getting the sum; Let: l= last term r= common ratio a= first term S= lr-a/ r-1 Now given these formulas, I have no idea how it is derived. I tried to formulate it on my own but I don't know how these equation was created. Please, give me tips to discern how a formula is made.
I agree! Memorizing simple formulas that are used often can be a time-saver, but one must also understand where the formulas come from, and if they are simple enough, there is no need to memorize at all; you can just derive them on demand. These formulas are a good example of this. First thing to recognize is that an arithmetical progression with first term = ##a## and common difference = ##d## simply looks like this: ##a, a+d, a+2d, a+3d, \ldots##. So the first term is ##a##, the second term is ##a+d##, the third term is ##a+2d##, and so forth. In general, the ##n##'th term is ##a+(n-1)d##. That's all there is to it. I see no point even memorizing such a simple thing. You can make a very similar argument for the geometric progression: the first term is ##a##, the second term is ##ar##, the third term is ##ar^2##, and in general, the ##n##'th term is ##ar^{n-1}##. The sum of the geometric progression is more interesting. Here, if you want to sum the first ##n## terms, you need to evaluate ##S = a + ar + ar^2 + \ldots + ar^{n-1}##. All of the terms have a common factor ##a##, so we can factor it out to get ##S = a(1 + r + r^2 + \ldots + r^{n-1})##. So the key is to find $$T = 1 + r + r^2 + \ldots + r^{n-1}$$ Note that we can multiply the entire equation by ##r## to obtain $$rT = r + r^2 + \ldots + r^{n-1} + r^n$$ Now (great trick) subtract the second equation from the first, to obtain $$T - rT = 1 - r^n$$ since all the other terms cancel. Now divide both sides by ##1-r## to get $$T = \frac{1 - r^n}{1-r}$$ Thus $$S = aT = a \frac{1 - r^n}{1-r}$$ is the answer you want. Note that it is valid as long as ##r \neq 1##. Can you find another formula for the ##r=1## case?
Hmmm... So I guess the book that I derived this from was just for introducing this concept. Its called Arithmetic for the practical man by Thomson. I never saw it that way, pretty interesting stuff. Thanks! What is the importance of series? or infinite series by the way? Sorry if I sound so ignorant since I really am. I heard its really important in higher mathematics.
When I first read your post I didn't notice you were also asking about the SUM of the geometric progression, which is more interesting. So I added some info about that to my previous response.
There are various reasons. One of the reasons is to be able to calculate and approximate functions. For example, if you want to calculate ##\sin(x)##, then expanding it in an infinite series will help. If you then take enough terms of the series, then you will have a real good approximation of the sine.
This is reason why mathematics is counter intuitively cool, I wish I know higher math right now, oh well.
By the way, the geometric series has (surprise!) a nice geometric interpretation, for example when ##r=1/2##. The sum ##S = 1/2 + 1/4 + 1/8 + 1/16 + \ldots## adds up to ##1## in the limit, because it is formed by adding half of the unit square (1/2), then half of the remaining area (1/4), then half of the remaining area (1/8), etc. See the pictures here: https://sites.google.com/site/butwhymath/m/geometric-series-visually
Yes, as I understand it, one form of Zeno's "paradox" is that since 1 can be expressed as 1/2 + 1/4 + 1/8 + 1/16 + ..., it's impossible to move 1 meter because it requires taking infinitely many smaller steps. What he didn't seem to consider is that the time required for each step is also shrinking geometrically, so the total time required is of course finite.
The easiest way for me is to visualize a circle. ##\frac{1}{2}## means 1 out of two.So you have total two parts and you select one from there.Similarly, if you have ##1 \over 100##.You have total 100 parts and you select 1 from there. Once you get the general idea,it will be as easy as pie.
Oh, of course, I know that visualization, but in some ways, for example, dividing a fraction by a fraction, I can't visualize it through a circle or a square but through a number line only. But I think I got the idea of adding a unit fraction by a unit fraction visually.
Dividing by a fraction is the same as multiplying by the reciprocal of the fraction. In this way you can turn any division problem into an equivalent multiplication problem. For example, instead of, say (1/2)/(1/3), this is the same as (1/2) * (3/1), or 3/2.
I didn't mean that. I meant to say, I can't visualize dividing a fraction by a fraction with the use of a figure, or shape, like circles or squares, but I can do the operation, abstractly, using numbers. Although I can picture it using the number line. And also, I can't visualize multiplying a fraction by a fraction using a model, using a figure, a shape, such as circle and square.
There's nothing wrong with that. The number line is a very good way to visualize division of fractions. If I have a line segment of length 1/2 and I want to subdivide it into line segments of length 1/8, how many do I need? Answer: 4. This is a perfectly good visualization of $$\frac{1/2}{1/8} = 4$$ In my opinion, trying to visualize it using squares or circles introduces unnecessary clutter.