# Arithmetic Series - Sum of n terms

## Arithmetic Series

A sequence is the set of the outputs of a function defined from the set of natural numbers to the set of real numbers or complex numbers. If the codomain of the function is the set of real numbers, it is called a real sequence and if it is the set of complex numbers on the other hand, it is called a complex sequence. For example: 1, 4, 9, 16, 25, 36, 49 ……….625 represents a sequence of squares of natural numbers till 25.

In very simple terms, a sequence is an ordered set of numbers. A sequence is denoted using braces. The sequencerepresents all the natural numbers. A sequence can be finite or infinite depending upon the number of terms it can have.

Now that we know what a sequence is let us learn about series. Sequence and series are very often confused with each other. Series are derived from sequences.

A series is defined as the sum of the terms of a sequence. It is denoted by

Where a_{i} is the i^{th} term of the sequence and I is a variable. ∑ is a symbol which stands for ‘summation’. It was invented by Leonard Euler, a Swiss mathematician.

The meaning of the above expression written using summation is:

Let us now discuss some special arithmetic series and their sum.

- 1 + 2 + 3 + 4 + ………. + n

This arithmetic series represents the sum of n natural numbers. Let us try to calculate the sum of this arithmetic series.

The difference between the sum of n natural numbers and sum of (n – 1) natural numbers is n, i.e.

S_{n} – S_{n-1} = n

Let us now proceed by taking the difference of sum of n natural numbers and sum of (n -2) natural numbers and so on,

S_{n} – S_{n-2} = n + (n – 1) = 2n -1

S_{n} – S_{n-3} = n + (n – 1) + (n – 2) = 3n – (1 + 2)

S_{n} – S_{n-4} = n + (n – 1) + (n – 2) + (n – 3) = 4n – (1 + 2 + 3)

Proceeding in the same manner, the general term can be expressed as:

According to the above equation the n^{th} term is clearly kn and the remaining terms are sum of natural numbers preceding it.

Now if k = n, equation 1 can be written as:

Since S_{0} is 0 the above sum becomes

Also sigma (p to n-1) {p} is one n short of S_{n}

Therefore,

Substituting the value of equation (3) in (2), we get

- 1
^{2}+ 2^{2}+ 3^{2}+ 4^{2}+ ………. + n^{2}

This arithmetic series represents the sum of squares of n natural numbers. Let us try to calculate the sum of this arithmetic series.

To prove this let us consider the identity p^{3} – (p – 1)^{3} = 3p^{2} – 3p + 1. In this identity let us put p = 1, 2, 3…. successively, we get

2^{3} – (1 – 1)^{3} = 3(1)^{2} – 3(1) + 1

2^{3} – (2 – 1)^{3} = 3(2)^{2} – 3(2) + 1

3^{3} – (3 – 1)^{3} = 3(3)^{2} – 3(3) + 1

………………………………………..

………………………………………..

………………………………………..

n^{3} – (n – 1)^{3} = 3(n)^{2} – 3(n) + 1

Adding both the sides of the equation, we get

We have already calculated the sum of n natural numbers as

Substituting this value n equation (4), we get

This represents the sum of squares of natural numbers using the summation notation. It can be simplified as:

- 1
^{3}+ 2^{3}+ 3^{3}+ 4^{3}+ ………. + n^{3}

This arithmetic series represents the sum of cubes of n natural numbers.

To prove this let us consider the identity (p + 1)^{4} – p^{4} =4p^{3} + 6p^{2} + 4p + 1. In this identity let us put p = 1, 2, 3…. successively, we get

2^{4} – 1^{4} =4(1)^{3} + 6(1)^{2} + 4(1) + 1

3^{4} – 2^{4} =4(2)^{3} + 6(2)^{2} + 4(2) + 1

4^{4} – 3^{4} =4(3)^{3} + 6(3)^{2} + 4(3) + 1

……..………………………………………..

………..……………………………………..

…………..…………………………………..

(n + 1)^{4} – n^{4} =4n^{3} + 6n^{2} + 4n + 1

Adding both the sides of the equation, we get

We have already calculated the sum of n natural numbers and sum of squares of n natural numbers as

Substituting these values in equation (4) and simplifying, we get

This represents the sum of squares of natural numbers using the summation notation. It can be simplified as: