Sequence

A list of numbers that are in order represent a sequence. For example, there is an odd numbers sequence {1,3,5,...}\{1, 3, 5, ...\} and the Fibonaccifibonacci sequence {0, 1, 1, 2, 3, 5, 8, 13, …}, and so on.

The order of a sequence can be increasing, decreasing, or a combination of both.

We can define a rule for any sequence. For example, identify the starting point of a sequence or the difference between the two consecutive numbers in a sequence.

Series

The sum of a sequence is called series. For example the odd series will be (1+3+5+7+9+...)(1+3+5+7+9+...)


Remember: The order matters in the case of a sequence, but a series is independent of the order. For example, 1,2,31, 2, 3 is not the same as 3,2,13, 2, 1 but 1+2+31+2+3 is just as same as 3+2+13+2+1. Similarly, the two sequences will have the same size if one sequence is the mirror image of the other. Like the following two sequences are of the same size (having log2n\log_2 n elements)

1,2,4,8,...,n/4,n/2,n1,2,4,8, ... , n/4, n/2, n

A geometric sequence with a common ratio of 22 and starting term is 11 and end term is nn.

n,n/2,n/4,...,8,4,2,1n, n/2, n/4, ..., 8, 4, 2, 1

A decreasing geometric sequence with common a ratio of 1/21/2 and starting term is nn and the end term is 11.


Arithmetic sequence

By establishing a rule (keeping the difference between two consecutive numbers the same in a sequence) the sequence will be converted to arithmetic sequence. Let’s say aa is the very first number of a sequence XX and dd is the difference between two consecutive numbers of that XX then any term in XX can be extracted from the following rule:

Xn=a+d(n1)X_n = a+d(n-1)

In simpler words, we can get nthn^{th} number of the sequence by adding the first term of the sequence with the product of common difference and n1n-1.

The common difference is not used in the first term. The first term is only aa and the next term comes by adding dd to it, and so on. So the general sequence should like this:

X=a+0d,a+d,a+2d,...,a+(n1)dX = a+0 \cdot d, a+d, a+2d , ... , a+(n-1)d

Arithmetic series (sum of natural numbers)

The arithmetic series is,

Sn=a+(a+d)+(a+2d)+...+(a+(n1)d)S_n = a+ (a+d) + (a+2d) + ... + (a+(n-1)d)

For example, for a=1a = 1 and d=1d=1, the above arithmetic sequence is nothing but the sum of the first NN positive integers.

Sn=1+2+3+4+...+nS_n = 1+2+3+4+...+n

The sum of first nn natural numbers can be calculated using this rule:

Sn=n2(n+1)S_n = \frac{n}{2}(n+1)

In order to understand why is it so, let’s calculate the sum of an arithmetic series (1+2+3+...+9+10)(1+2+3+...+9+10). According to the given formula:

S1...10=102(10+1)=55S_{1...10} = \frac{10}{2}(10+1)=55

We get the sum to be 55, which is true, by adding the numbers from 11 to 1010 on each slide. Let’s see if we can figure out how it’s done!

Firstly, in the above slides, each element of the series was paired. 1 is paired up with the last number 10, 2 is paired with the second last number 9 yielding 11 again, so a total of 10/2=510/2=5 pairs are created, where the sum of each pair is 1111. Hence, the total sum will be 10/5×11=5510/5\times11=55.

Generally, for the sum of the first nn numbers, 11 is paired with nn yielding 1+n{1+n}, 2 is paired with n1n-1 yielding 2+(n1)=n+12+(n-1)=n+1, and so on. So a total of n2\frac{n}{2} pairs will be created, where the sum of each pair is (n+1)(n+1). Hence, the total sum will be Sn=n2(n+1)S_n = \frac{n}{2}(n+1).

Generalized arithmetic series

Question

What will be the sum of the following general arithmetic series: Sn=a+a+d+a+2d+...+a+(n1)dS_n = a+ a+d + a+2d + ... + a+(n-1)d

Show Answer

Application of arithmetic series

Suppose we have an infinite number of planes and lines, lines can intersect each other to make regions, and our objective is to maximize the number of regions. Can we come up with a rule to find out the maximum number of regions for nn number of lines in a plane?

Directions

With no line drawn, the entire plane is just one region. By adding a line, a new region will be produced. Now, when we’ll add a new line, 22 new regions will be created. Upon adding the third line, 33 new regions will be created.

Solution

First of all, let’s visualize the pattern in the following slides.

From the above, we can determine that 11 line produces 22 regions, 22 lines produce 44 regions. With three lines we can have at most 7 regions. For nn number of lines the total number of regions will be as follows:

Number of lines New regions Total number of regions
00 - 11
11 11 1+(1)=21+(1)=2
22 22 1+(1+2)=41+(1+2)=4
33 33 1+(1+2+3)=71+(1+2+3)=7
44 44 1+(1+2+3+4)=111+(1+2+3+4)=11
\vdots \vdots \vdots
n n ???

Two things to note here:

  1. We have 11 region before any line, such asthe entire plane.
  2. For maximizing the number of regions, each ii'th newly added line should create the maximum number of new regions which is only possible if the ii'th line not only intersects with every already present line that is i1i-1 previous line but also at a different point of intersection. Hence, creating ii new regions (i1i-1 for intersecting every line and +1 for dissecting the final region from where the line is exiting into the infinite space). Therefore, the total number of regions with nn lines are as follows:
    •  1+\space1+ (initial empty infinite plane)
    •    +1\space \space \space+1 (for the first line)
    •    +2\space \space \space+2 (for the second line)
    •    +3\space \space \space+3 (for the third line)
    •    +...\space \space \space+ ...
    •    +n\space \space \space+ n (for the nn'th line)
    • =1+n×(n+1)2= 1+ \frac{n \times (n+1)}{2}

Exercise 1: Counting handshakes

On your first day at college, your social science teacher suggests that it would be a good idea for each student to meet every other student in the class. If there were 2020 students in the class,

  1. How many total handshakes would happen?
  2. What if there were NN students? Can you tell the total number of handshakes in terms of NN?

Directions to look at the exercise

One student will shake hands with another student only once. For example, if we have A, B, and C, then A will shake hands with both B and C. But B will only shake hands with C because he has already shaken hands with A. Now, C does not need to shake hands with A or B because he has already shaken hands with them.

Exercise 2: Counting squares in a chess board

If we have an 8 x 8 chessboard, how many squares (within that chessboard) would there be in total? (64 is not the answer).