Series Part 2

Continuing from the previous post on this subject, lets find the formulas for the sum of the two series I mentioned in the previous post, the arithmetic and the geometric progressions:

Arithmetic progression:
Firstly, lets look on a very simple case - the progression 1+2+3+...+n=S. If we just look on it, it is not very clear what the sum is. However, if we will write it in a slightly different manner the answer will be obvious:

1,2,3,4,5,6,7,8,9,...n
n...,9,8,7,6,5,4,3,2,1

Obviously the sum of all the numbers is 2S. We also know that there are n columns and the sum of any column is n+1. Therefore 2S=n(n+1). From this we get the formula by division.

Now lets look on the general case. The general arithmetic series is:

a+(a+d)+(a+2d)+...+(a+(n-1)d)=S
na+d+2d+3d+...+(n-1)d=S
na+d(1+2+...+(n-1))=S
na+0.5d(n-1)n=S
0.5[2a+(n-1)d]n=S

By the way, this formula requires to know the first number in the progression. However, if you know the last one instead of the first you can also use this formula - after changing d to (-d).


Geometric progression:
Firstly, lets again consider a specific case. Lets look on the sum of the following progression:

1+q+q^2+q^3+..+q^(n-1)=S
q+q^2+q^3+..+q^(n-1)+q^n=Sq
q^n-1=Sq-S
S=(q^n-1)/(q-1)

Getting from this to the general case is extremely easy. The only thing we need to do is to remember that a general geometric progression is just a multiplication of the case we took care of by a constant number a. Therefore, the formula for the sum is:

a+aq+aq^2+....+aq^(n-1)=S
S=a(q^n-1)/(q-1)

In the next post I will talk about infinite series and their sums.