Special Series in Maths Sequences and Series | Class 11 Maths (original) (raw)

Last Updated : 23 Jul, 2025

**Special Series: A series can be defined as the sum of all the numbers of the given sequence. The sequences are finite as well as infinite. In the same way, the series can also be finite or infinite. For example, consider a sequence as **1, 3, 5, 7, … Then the series of these terms will be **1 + 3 + 5 + 7 + ….. The series special in some way or the other is called a special series. The following are the **three types of special series.

  1. **1 + 2 + 3 +… + n (sum of first n natural numbers)
  2. **1 2 + 2 2 + 3 2 +… + n 2 (sum of squares of the first n natural numbers)
  3. **1 3 + 2 3 + 3 3 +… + n 3 ****(sum of cubes of the first n natural numbers)**

In this article, we will read about special series in maths and how to get the formula for all these series. We will also see solved examples and practice problems on special series.

Table of Content

What are Special Series?

Special series in mathematics refers to specific sequences of numbers or terms that follow a particular pattern or rule and are widely studied for their unique properties and applications. These series include well-known examples like the arithmetic series, geometric series, harmonic series, and Fibonacci series.

Special Series 1: S**um of first n natural numbers

The result of this series is given below:

**1+ 2 + 3 + 4 + .... + n = n (n + 1) / 2

**Proof:

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

We can see that this is an Arithmetic Progression with the first term (a) = 1 and common difference (d) =1 and there are n term

So, Sum of n terms = n/2 (2 x a + (n - 1) x d)

Putting the values for this series we will get

Sn = n/2(2 x 1 + (n - 1) x 1)

Sn = n/2(2 + n - 1)

Sn = n(n + 1)/2

**Hence Proved.

Example on Sum of first n natural numbers

**Question. Find the sum of the following series 3 + 4 + 5 ---- + 25?

**Solution:

Let Sn = 3+ 4 + 5 -- + 25

Now we can also write it like this

Sn + 1 + 2 = 1 + 2 + 3 + 4 ---- + 25

Clearly now it is the sum of first 25 natural number we can be written like this

Sn + 1+ 2 = 25 (25 + 1) / 2

Sn = 325 - 1 - 2

Sn = 322

Special Series 2: S**um of squares of the first n natural numbers

The Result of this series is given below:

**1 2 + 2 2 + 3 2 +… + n 2 **= n(n + 1) (2n + 1)/6

**Proof:

Let Sn = 12 + 22 + 32 +… + n2 **---eq 1

We know that, k3 – (k – 1)3 = 3k2 – 3k + 1 **--- eq 2

We know that, (a - b)3 = a3 - b3 - 3a2b + 3ab2

So, k3 - (k - 1)3

= k3 - k3 +1 + 3k2 - 3k

= 3k2 - 3k +1

Putting k = 1, 2…, n successively in **eq 2, we obtain

13 – 03 = 3(1)2 – 3(1) + 1

23 – 13 = 3(2)2 – 3(2) + 1

33 – 23 = 3(3)2 – 3(3) + 1

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n3 – (n – 1)3= 3(n)2 – 3(n) + 1

Adding both sides of all above equations, we get

n3 – 03 = 3 (12 + 22 + 32 + ... + n2) – 3 (1 + 2 + 3 + ... + n) + n

We can write this like:

n3 = 3 ∑(k2) - 3∑(k) +n, where 1 ≤ k ≤ n **--- eq(3)

We know that,

**∑(k) (where 1 ≤ k ≤ n ) = 1 + 2 + 3 + 4 --- n = n(n + 1)/2 ---eq(4)

and **eq 1 can also be written like this

Sn = ∑(k2), where 1 ≤ k ≤ n **--- eq(1)

Now, putting these values in **eq 3

n3 = 3Sn - 3(n)(n + 1)/ 2 + n

n3 + 3 (n) (n + 1)/2 - n = 3Sn

(2n3 + 3n2 + 3n - 2n)/2 = 3Sn

(2n3 + 3n2 + n)/6 = Sn

n(2n2 + 3n + 1)/6 = Sn

n(2n2 + n + 2n + 1)/6 = Sn

n(n(2n + 1) + 1(2n + 1))/6 = Sn

n(n + 1)(2n + 1)/6 = Sn

Sn = n(n + 1)(2n + 1)/6

**Hence proved.

Examples on Sum of squares of the first n natural numbers

**Question 1. Find the sum of the n terms of the series whose nth terms is n 2 **+ n + 1?

**Solution:

Given that ,

an = n2 + n + 1

Thus, the sum to n terms is given by

Sn = ∑ak ****(where 1 ≤ k ≤ n )** = ∑ k2 + ∑ k + ∑1 ****(where 1 ≤ k ≤ n)**

= n(n + 1) (2n + 1)/6 + n (n + 1)/2 + n

= (n(n + 1) (2n + 1) + 3n(n + 1) + 6n)/6

= ((n2+ n) (2n + 1) + 3n2 + 3n + 6n)/6

= (2n3 + 2n2 + n2 + n + 3n2 + 9n)/6

= (2n3 + 6n2 + 10n)/6

**Question 2. Find the sum of the following series up to n terms 1 + 1 + 2 + 1 + 2 + 3 + 1 + 2 + 3 + 4 + ------?

**Solution:

If we observe the series carefully we can write it like this

**S n =(1) + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ------

We can say that we have to find sum of the sum of first n natural number.

So we can write Sn= **Σ((i(i + 1))/2), where 1 ≤ i ≤ n

= (1/2)Σ (i(i + 1))

= (1/2)Σ (i2 + i)

= (1/2)(Σ i2 + Σ i)

We know Σ i2 = n (n + 1) (2n + 1) / 6 and

Σ i = n (n + 1) / 2.

Substituting the value, we get,

Sum = (1/2)((n(n + 1)(2n + 1) / 6) + (n( n + 1) / 2))

= n(n + 1)/2 [(2n + 1)/6 + 1/2]

**= n(n + 1)(n + 2) / 6

Special Series 3: S**um of cubes of the first n natural numbers

The Result of this series is given below:

**1 3 + 2 3 + 3 3 + … + n 3 = (n (n + 1)/2) 2

**Proof:

Let Sn = 13 + 23 + 33 +… + n3 **---eq 1

We know that, (k + 1)4 – (k)4 = 4k3 + 6k2 + 4k + 1 **--- eq 2

We know that, (a+b)4 = (a2 +b2 +2ab)2

= a4 + b4 + 6a2b2 + 4a3b + 4ab3

So, (k + 1)4 - (k)4

= k4 + 1 + 6k2 + 4k3 + 4k- k4

= 4k3 +6k2 + 4k +1

Putting k = 1, 2…, n successively in **eq 2 , we obtain

(1 + 1)4 – 14 = 4(1)3 + 6(1)2 + 4(1) + 1

(2 + 1)4 – 24 = 4(2)3 + 6(2)2 + 4(2) + 1

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(n + 1)4 – (n)4 = 4(n)3 + 6n2 + 4n + 1

Adding both sides of all the above equations, we get

(n + 1)4 – 14 = 4 (13 + 23+ 33 + ... + n3) + 6(12 + 22+ 32 + 42 + 52) + 4 (1 + 2 + 3 + ... + n) + n

We can write this like:

(n + 1)4 - 14 = 4 ∑ (k3) + 6∑(k2) + 4∑(k) + n where 1 ≤ k ≤ n **--- eq(3)

We know that ,

∑(k) (where 1 ≤ k ≤ n ) = 1 + 2 + 3 + 4 --- n = n (n + 1)/2 **---eq(4)

∑(k2) (where 1 ≤ k ≤ n ) = 12 + 22 + 32 + 42 --- n2 = n (n + 1) (2n + 1)/6 **---eq(5)

and **eq 1 can also be written like this

Sn = ∑(k3) , where 1 ≤ k ≤ n --- eq(1)

Now, putting these values in **eq 3

(n + 1)4 -14 = 4Sn+ 6(n) (n + 1) (2n + 1)/6 + 4 (n) (n + 1)/2 + n

n4 + 6n2 + 4n3 + 4n - (n)(2n2 + 3n + 1) - 2(n)(n + 1) - n = 4Sn

n4 + 6n2 + 4n3 + 4n - 2n3 - 3n2 - n - 2n2 - 2n - n = 4Sn

n4 + n2 + 2n3 = 4Sn

n2 (n2 + 1 + 2n) = 4Sn

n2 (n + 1)2 = 4Sn

Sn = (n(n + 1)/2)2

**Hence proved.

Example on Sum of cubes of the first n natural numbers

**Question. Find the value of the following fraction (1 3 + 2 3 + 3 3 ---- + 9 3 ) / (1 + 2 + 3 ---- + 9)?

**Solution:

**Sum of first n natural number : n(n + 1)/2

**Sum of cube of first n natural number : (n(n + 1)/2)2

So, (13 + 23 + 33 ----+ n3) / (1+ 2+ 3 ---- +n)

= ((n(n + 1)/2)2) / (n(n + 1)/2)

= n(n + 1)/2

Now as we can see that value of n is 9 in the question,

= 9 (9 + 1) / 2

= 9 x 5

= 45

Practice Problems on Special Series

**1. Find the sum of the first 20 terms of the arithmetic series where the first term is 5 and the common difference is 3.

**2. Calculate the sum of the first 8 terms of the geometric series where the first term is 2 and the common ratio is 3.

**3. Find the sum of the first 5 terms of the harmonic series.

**4. Find the 10th term in the Fibonacci series.