Thread:Light and Bright/@comment-35776050-20190331132944/@comment-36247755-20190331231914

Part 2:

Formulas

Arithmetic and geometric sequences have formulas so that you can find any term in the sequence.

Here are the formulae regarding arithmetic sequences: un = a + [(n - 1) × d] Sn = n[2a + (n - 1) × d]/2
 * Finding the nth term:
 * Finding the sum of the first n terms:

 a = the first term in the sequence

n = the term number (1st term, 2nd term, 3rd term, etc.)

d = the common difference (difference between consecutive terms) 

''' What are these formulae and why do they work? '''

Let's start with the first formula. un = a + [(n - 1) × d]

We'll try it out with an example, say this sequence:

The first part of the formula has "a," which makes sense, it being the first term.

The second part gets a little trickier. [(n - 1) × d]

For the first term, the formula results in "a," which is expected.

For the second term, the formula results in "a + d." This signifies that the second term has the common difference added to it.

For the third term, the formula results in "a + 2d." Another common difference was added to the previous term.

This is why the formula works; it starts with the first term, and has the common difference added to it.

Now for the sum formula.

To know how this works, one must understand Gaussian addition. You basically take the pairs of numbers from the ends inwards.



This is exactly how the sum formula works. If you were to split apart the formula, you'd get the following:

Sn=n(a + [a + (n - 1) × d])/2

Now you can see how we are adding the first term and the last term, and multiplying it by the number of number pairs.

 Geometric sequence formulae 

Hopefully I won't need to explain the first formula, since pretty much the same explanations go for it. The second formula is more complicated. If you really want me to, I can explain it, but you probably won't need it for your test tomorrow. The third one should be self-explanatory once you try out an example or two.

gn = a × r(n - 1) Sn = a[(1 - rn)/(1 - r)] Sn = a/(1 - r)
 * Finding the nth term:
 * Finding the sum of the first n terms:
 * Decreasing infinite geometric series formula:

 a = the first term in the sequence

n = the term number (1st term, 2nd term, 3rd term, etc.)

r = the common ratio (ratio between consecutive terms) 