A **sequence** is said to be in a progression if its last term can be represented using a formula. If the co-domain 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.

3, 7, 11, 15, 19,………..87 forms another sequence, where each of the term exceeds the preceeding term by 4.

If all the terms of a progression except the first one exceeds the preceding term by a fixed number, then the progression is called **arithmetic progression**. If \(a\)

\(a, a~+~d, a~+~2d, ………, a~+~(n-1)d\)

Before learning how to derive a formula to get the sum of n terms in an AP, try this activity:

Try to get the sum of first 100 natural numbers without using any formula.

This question was posed in the same way to one of the great mathematicians, Carl Gauss (1777-1855). He is often referred to be ‘Princeps *mathematicorum*’ (Latin), meaning ‘the foremost of mathematicians’. At that time, his age was 10 yrs. He came up with the answer of the above problem in matter of seconds.

The proof for the question can be done using the following way:

The sum of number can be represented as

\(Sum = 1+2+3+4+……………+ 97 + 98 + 99 + 100\)

Even if the order of the numbers is reversed, their sum remains same.

\(Sum = 100 + 99 + 98 + 97 + ………..+ 4 + 3 + 2 + 1\)

Adding equations 1 and 2, we get

\(2 \times Sum = (100+ 1) + (99+2) + (98+3 )+ (97 +4)+ ………..(4+97)+(3+98)+(2+99)+(1+100)\)

\(\Rightarrow 2 \times Sum = 101 + 101 + 101 + 101 + ………..(4+97)+(3+98)+(2+99)+(1+100)\)

\(\Rightarrow 2 \times Sum = 101 (1 + 1 + 1 + …..100 \;\; terms)\)

\(\Rightarrow 2 \times Sum = 101 (100)\)

\(\Rightarrow Sum = \frac{101 \times 100}{2}\)

\(\Rightarrow Sum = 5050\)

Using the above method, sum of numbers like 1000, 10000, etc. can also be calculated.

Table 1: Table showing the sum of natural numbers

Numbers |
Sum |

1-10 | 55 |

1-100 | 5050 |

1-1000 | 500500 |

1-10000 | 50005000 |

1-100000 | 5000050000 |

1-1000000 | 500000500000 |

The interesting thing is that the above method is applicable to any AP (if the last term of the AP is known). Consider the general form of AP with first term as a, common difference as \(d\)

*\(Sum\)* = \(a~+~(a~+~d)~+~(a~+~2d)……+~(l~-~2d)~+~(l~-~d)~+~l——————– (3)\)

where \(l\)

Writing in reverse order, the sum will still remain same.

*\(Sum\)* = \(l~+~(l~-~d)~+~(l~-~2d)..…~+~(a~+~2d)~+~(a~+~d)~+~a~——————- (4)\)

Adding equations 3 and 4, we get

\(2\)*Sum* = \((a~+~l)~+~[(a~+~d)~+~(l~-~d)]………~+~[(l~-~d)~+(a~+~d)]~+~(l~+~a)]\)

\(⇒~2× \)*Sum* = \((a~+~l)~+~(a~+~l)………+~(a~+~l)~+~(a~+~l)\)

\(⇒~2\)*Sum* = \(n×(a~+~l)\)

**⇒ Sum = \(\frac{n}{2}(a~+~l)\)**

Substituting the value of \(l\)

**\(Sum ~of~ n ~terms~ of~ AP\) = \(\frac{n}{2}[2a~+~(n~-~1)d]\)**

For AP of natural numbers**,** **a = 1** and **d = 1, **Sum of \(n\)

\(S_n\)

\(\large \mathbf{\Rightarrow S_{n} = \frac{n(n+1)}{2}}\)

Some examples will enhance the understanding of the topic.

*Example 1*: If first term of an AP is 67 and common difference is -13, find the sum of first 20 terms.

** Solution:** Here, \(a\)

\(S_n\)

\(S_{20}\)

\(S_{20}\)

\(S_{20}\)

So, the sum of first 20 terms is \(-1130\)

*Example 2*: The sum of \(n\)

** Solution:** The sum of \(n\)

Similarly, \(S_{n~-~1}\)

\(a\)

\(d\)

For an AP, \(S_n\)

Putting \(n\)

\(l\)

\(S_n~-~S_{n~-~1}\)

So, \(441~-~356\)

\(a_n\)

Since \(d\)

\(n(n~-~1)\)

\(⇒~n^2~-~n~-~72\)

Solving by factorization method,

\(n^2~-~9n~+~8n~-~72\)

\((n~-~9)(n~+~8)\)

So, \(n\)

Since number of terms can’t be negative,

\(n\)

**Sum of Square Series**

**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

\(\large n^{3}- 0^{3} = 3(1^{2}+ 2^{2}+ 3^{2}+……+n^{2}) – 3 (1+ 2+ 3+ ……+ n) + n\)

\(\large \Rightarrow n^{3} = 3\sum_{k=1}^{n}(k^{2}) – 3 \sum_{k=1}^{n} (k) + n\)

We have already calculated the sum of n natural numbers as

\(\large \sum_{p=1}^{n-1}p = S_{n} = \frac{n(n+1)}{2}\)

Substituting this value n equation (5), we get

\(\large \sum_{p=1}^{n-1}p^{2} = n^{3} – 3 \left (\frac{n(n+1)}{2} \right ) + n\)

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

\(\large S_{n} = \sum_{p=1}^{n-1}p^{2} = \frac{1}{3}\left [ n^{3}+ \frac{3n(n+1)}{2} – n \right ] = \frac{1}{6} \left [ 2n^{3} + 3n(n+1) – 2n \right ] = \frac{n(n+1)(2n+1)}{6}\)

**Sum of Cubic Series**

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

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

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

\(\large (n+1)^{4} – 1^{4} = 4 (1^{3} + 2^{3} + 3^{3} + …..+ n^{3}) + 6 (1^{2} + 2^{2} + 3^{2} + …..+ n^{2}) + 4 (1 + 2+3+4+…..+ n) + n\)

\(\large \Rightarrow n^{3} = 4 \sum_{p = 1}^{n}p^{3} + 6 \sum_{p = 1}^{n}p^{2} + 6 \sum_{p = 1}^{n}p^{p} + n\)

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

\(\large \sum_{p = 1}^{n}p = S_{n} = \frac{n(n+1)}{2}\)

Substituting these values in equation (5) and simplifying, we get

\(\large 4\sum_{p = 1}^{n}p^{3} = n^{4} +4n^{2} + 6n^{2} + 4n – 6\left (\frac{n(n+1)(2n+1)}{6} \right ) -4\left (\frac{n(n+1)}{2} \right ) + n\)

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

\(\large S_{n} = \sum_{p=1}^{n} p^{3} = \frac{1}{4} \left [ n^{2} (n+1)^{2} \right ] = \left (\frac{n(n+1)}{2} \right )^{2}\)

To practice more questions on sum of n terms of the topic, download BYJU’S – The Learning App from Google Play Store.

**Practise This Question**