A number pattern is a unique quality that prevails in a set of numbers. More than 200 years back, an elementary school teacher in Germany wanted to engage the students of his class for some time by asking them to find the sum of all the numbers from 1 to 100. But one student amazed the teacher by telling the correct answer almost instantly. Because he found a pattern that every pair of numbers that are in the same place from either ends adds up to the same sum of 101. Hence the total is 50 pair’s times 101 = 5050. The student was none other than the great mathematician and scientist by his well-known name ‘Gauss’. Thus the existence of number pattern was noticed from ancient times.

A list of numbers which form a pattern is called a sequence. The arithmetic and geometric sequences developed in Algebra are based on number patterns. The set of numbers 1, 2, 3, 4 … exhibit a pattern that the difference between any two successive terms is constant and that is equal to 1. Such a sequence with a common difference (not necessarily 1 and could be any integer) is called an arithmetic sequence.
In a geometric sequence, the pattern is the common ratio that exists between consecutive terms. By applying the concept of number patterns, you can predict any term in a sequence and also predict the sum of finite number of terms. In fact, in a geometric sequence, if the common ratio is a proper fraction, even the sum of infinite terms could be figured out.

Finding Number Patterns

The question ‘how to do number patterns means ‘how do you find the pattern’ in a set of numbers. All the sequences of numbers may not exhibit an arithmetic pattern or a geometric pattern. The pattern may also be ‘hidden’, meaning that at the first cite you may not recognize the pattern. Let us see the following sequence as an example.
1, 3, 6, 10 …

At the outset we can conclude that the sequence is neither arithmetic nor geometric. The reason is, the difference between two successive terms is not constant and also the ratio between two successive terms is also not a constant. But let us study in detail as what is ‘hidden’ here. Let us rewrite the sequence in a different form.
1, 3, 6, 10, 15 … = 1, (2 + 1), (3 + 3), (4 + 6), (5 + 10)…

So we see that any term is the sum of that term number and the previous term.
Next, let us consider the sequence 12, 22, 32, 42 … One can easily notice that the pattern is, the number in any term is the square of that term number. But what is more interesting is this sequence helps to find another number pattern. Let us see what it is.
12 = 1
22 = 4 = 1+ 3
32 = 9 = 1 + 3 + 5
42 = 16 = 1 + 3 + 5 + 7
So, the sum of ‘n’ odd integers is n2!
We will see yet another interesting sequence.
1, 1, 2, 3, 5, 8, 13…

Can you guess what the pattern here is? Yes, the number in any term is the sum of the numbers of preceding two terms. The numbers of this sequence are called as Fibonacci numbers.

Thus, there are innumerable number patterns in the whole number systems. The concept of number patterns is not learned for fun. Many number patterns have practical use.

For example, Fibonacci numbers are seen in various plant growths and the coefficients in binomial expansions follow a triangular pattern.

Number Pattern Problems

Since number patterns have many practical usages, the usages are figured out by solving problems on number pattern. Broadly, the number pattern problems are based on finding a particular term or finding the sum of a finite number of terms of a given sequence. The first task is to identify the pattern in the given sequence. Once you determine that you can form recursive formula for that particular sequence.

Solving Number Patterns

Solved Example

Question: Find a recursive formula and find the 6th term of the following sequence. 2, 6, 15, 34…
Solution:

Since no pattern is apparently seen, let us see whether the term numbers could help. So let us split each term as a sum of two numbers, the second one as the term number. So,
2 = (1 + 1), 6 = (4 + 2), 15 = (12 + 3), 34 = (30 + 4)
In all the decompositions, the first addend is twice the number in preceding term. Therefore, if an is the number in any term ‘n’, the recursive formula for this sequence is,
an = 2an-1 + n, with first term as 2.
Now, to find the 6th term, we should first find the 5th term.
a5 = 2a4+ 5 =  2 * 34 + 5 = 73 and hence,
a6 = 2a5+ 6 =  2 * 73 + 6 = 152.