1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987

(note: some list "0" as the first term others start with the "1" as the first term)

i.e. im looking for an equation where i punch in the term number (n) and it spits out the value (i.e. i input "7" for n and it gives me 13)

t_n = n(n+1)/2

(^this is obviously NOT the right equation)

im not entirely sure if one exists, i heard there is one that involves cosines and stuff, but im thinking there just isn't an equation for it without using Loops in a computer program to find out the value

thanks in advance for the help

Recursive equations tend to lead to power functions as solutions so a good way to approach almost any such equation is:
Look for a solution of the form f_n= axⁿ.

The fibonacci numbers satisfy f_{n + 2}= f_{n + 1} + f_n or a x^{n + 2}= ax^{n + 1} + axⁿ. We can divide by a and rewrite as x^{n + 2} - xⁿ⁺¹ - xⁿ= 0. x= 0 will satisfy the recursive equation but not the initial requirements that f₀= f₁= 1 so we can divide through by x to get x²- x- 1= 0. There are, of course, two roots to that equation: x= (1+5)/2 and x= (1-5)/2.

As is usual with such linear problems, the general solution is a linear combination of these: f_n= A((1+5)/2)ⁿ+ B((1-5)/2)ⁿ.

Solve for A and B so that f₀= f₁= 1.

Note that this formula gives the sequence as F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, etc... Replace both n's in the formula with the term number, and it will give you the corresponding value.

"How it is done" is much too involved to post here. If you want to learn how to find explicit formulas for sequences like the fibonacci numbers, the topic to look up is "solving linear homogeneous recurrence relations with constant coefficients". This is normally covered in courses on discrete mathematics. Here are urls for a couple of sites that seem to cover this topic fairly well:

http://people.uncw.edu/tompkinsj/133/recursion/homogeneous.htm
http://www.cs.princeton.edu/courses/archive/fall02/cs341/lec9-10.pdf

I thought the "/" was one big square root symbol, instead of divided by the square root

ie f_n=(1/5)(ⁿ - 1/ⁿ)