An oversimplification of sequences and summations may be useful to gain a basic understanding of these two concepts. Both sequences and summations are strings of elements (representing numbers in this discussion) where successive elements are separated either by a comma (in the case of sequences) or a plus sign (in the case of summations).

1. 2, 3, 5, 7, 11 is a sequence of 5 positive integers
2. 1, 3, 5, 7, 9, …, 2*n-1 is a sequence of n positive integers

1. 2 + 3 + 5 + 7 + 11 is a summation of 5 positive integers.
2. 1 + 3 + 5 + 7 + 9, …+ 2*n-1 is a summation of n positive integers

This may also be written as å (2*i-1) where (1<=i<=n)

Geometric and Arithmetic Summations:

There are two basic types of summations, arithmetic and geometric. In each, any term may be computed from the previous term by either adding a constant (as in the case of arithmetic summations) or by multiplying by a constant (as in the case of geometric summations). For example, in the arithmetic summation 1+3+5+7+9+…+2*n-1, each term is 2 greater than the previous term.

Similarly in the geometric summation 1/2 + 1/4 + 1/8 + 1/16 + … + 1/2^N

Each term may be computed by multiplying the previous term by 1/2.

Many formulas exist for arithmetic and geometric summations.
Consider a finite geometric summation where the first term is a, the ratio between successive terms is r and the number of terms is n. The formula for the sum of the first n terms of such a geometric summation is
(a - a*rⁿ⁺¹)/(1-r).

If the ratio, r, between successive terms is less than 1 then the sum is finite even if the number of terms is infinite. The formula for the sum of an infinite geometric summation is simply a/(1-r).

The formula for an arithmetic summation is not given in your text but we covered it in class. Roughly stated

Sum = (First Term + Last Term) * (Number of Terms)/2

The (Number of Terms)/2 is the "Number of Pairs" which we discussed in class. I have stated it in the formula as (Number of Terms)/2 to emphasize that the formula holds true even when the number of terms is odd; thus removing from consideration the idea of a half of a pair.

1. Compute the following sum: å (2*k - 1) (1<=k<=N) = 1 + 3 + 5 + … + 2*N – 1

This we recognize to be an arithmetic summation. The first term is 1. The last term is 2*N-1 and the number of terms is N. Therefore

Sum = (1 + 2*N - 1) * N/2 = N²

 

2. Compute the following sum: å 1/3^K (1 <= K <= 5) = 1/3 + 1/9 + 1/27 + 1/81 + 1/243

Note that arithmetic and geometric summations are also referred to as arithmetic and geometric series.

Double summations may be evaluated using a two step process.

1. First expand the inner summation and
2. Then, expand the outer one.

Expand the following summation: å _I²₌₁ å _J³_{= 1}(I + J)

å _I²₌₁ å _J³_{= 1}(I + J) =

å _I²₌₁ (I + 1) + (I + 2) + (I + 3) =

å _I²₌₁ (3*I + 6) =

(3*1 + 6) + ( 3*2 + 6) = 9 + 12 = 21