Maths n00b

[Naureg]

For a sequence like this:

2 4 6 8 16 32

or

500 1000 2000 4000

(Where the previous term is doubled)

The equation is:

a = base
r = multiplier (doubling/tripling previous term)
n = term number, as usual

x = a(rⁿ)

I rearranged this to get:

n = log_r(x/a)

Since I'm only in high school I'd never done logarithms so the above was hard to rearrange ^_^

What I want to know is for the same sequence but each term is added up, what is the equation?

And what would that be rearranged to give n?

[atropos]

What I want to know is for the same sequence but each term is added up, what is the equation

How do you mean? Could you give an example of the sequence you have in mind?

[chrisname]

A term in a geometric series is given by the equation U_n = ar^{n - 1}, so the value of n is log_r(U_n/a) + log_rr = log_r(U_n/a) + 1.

The other one you're thinking of is called an arithmetic series and the equation is U_n = a + (n - 1)d, so (n - 1)d = -a, so n - 1 = -a/d, so n = (-a/d) + 1.

I think I've rearranged that correctly, it's easier on paper (so you might want to do it yourself to check; the equation a + (n - 1)d is right though).

[Gaminic]

There is no equation for the sum, in a general case. If you have a finite set of numbers, simply sum them. If you want the infinite/asymptotic sum, you'll have to check for convergence (which in this case is false, thus the sum would be infinity).

[ne555]

However, for a geometric serie the sum is simply
\sum_{K=0}^{n-1} r^{K-1} = \frac{1-r^n}{1-r} \qquad \forall r \neq 1

[chrisname]

If a geometric series is infinite and r < 1, then the sum is http://texify.com/?$\frac{a}{1 - r}$
If it's finite, the sum is http://texify.com/?$a\frac{1 - r^n}{1 - r}$

For an arithmetic series, the sum is http://texify.com/?$\frac{n}{2}(a_1 + a_n) = \frac{n}{2}[a_1 + (n - 1)d]$

(copy the whole link (up to and including the $) to see the equation).