Special Series in Maths – Sequences and Series | Class 11 Maths

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. 1² + 2² + 3² +… + n² (sum of squares of the first n natural numbers)
3. 1³ + 2³ + 3³ +… + n³ (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.

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.

The result of this series is given below:

1+ 2 + 3 + 4 + …. + n = n (n + 1) / 2

Proof:

Let S_n = 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 

S_n = n/2(2 x 1 + (n – 1) x 1)

S_n = n/2(2 + n – 1)

S_n = 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 S_n = 3+ 4 + 5 — + 25

Now we can also write it like this 

S_n + 1 + 2 = 1 + 2 + 3 + 4 —- + 25

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

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

S_n = 325 – 1 – 2

S_n = 322 

The Result of this series is given below:

1² + 2² + 3² +… + n²  = n(n + 1) (2n + 1)/6

Proof:

Let S_n = 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

…………………………………

…………………………………

………………………………..

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

S_n = ∑(k²), where 1 ≤ k ≤ n  — eq(1)

Now, putting these values in eq 3

n³ = 3S_n – 3(n)(n + 1)/ 2 + n

n³ + 3 (n) (n + 1)/2 – n = 3S_n

(2n³ + 3n² + 3n – 2n)/2 = 3S_n

(2n³ + 3n² + n)/6 = S_n

n(2n² + 3n + 1)/6 = S_n

n(2n² + n + 2n + 1)/6 = S_n

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

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

S_n = 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 n² + n + 1?

Solution: 

Given that , 

a_n = n² + n + 1

Thus, the sum to n terms is given by

S_n = ∑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 S_n=  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

The Result of this series is given below:

1³ + 2³ + 3³ + … + n³  = (n (n + 1)/2)²

Proof:

Let S_n = 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

…………………………………

…………………………………

………………………………..

(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

S_n = ∑(k³) , where 1 ≤ k ≤ n  — eq(1)

Now, putting these values in eq 3

(n + 1)⁴ -1⁴ = 4S_n+ 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 = 4S_n

n⁴ + 6n² + 4n³ + 4n – 2n³ – 3n² – n – 2n² – 2n – n = 4S_n

n⁴ + n² + 2n³ = 4S_n 

n² (n² + 1 + 2n) = 4S_n

n² (n + 1)² = 4S_n

S_n = (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 (1³ + 2³ + 3³ —- + 9³) / (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

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.

What is an arithmetic series and how do you find its sum?

An arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. To find the sum of the first nnn terms of an arithmetic series, you can use the formula: S_n = n²(2a + (n – 1)d)

How do you calculate the sum of a geometric series?

A geometric series is a sequence of numbers where each term is multiplied by a constant ratio to get the next term. The sum of the first nnn terms of a geometric series can be calculated using the formula:

[Tex]S_n = a \frac{r^n – 1}{r – 1} [/Tex]

What is a harmonic series and where is it used?

A harmonic series is a sequence of numbers where each term is the reciprocal of an integer. The nnnth harmonic number is the sum of the reciprocals of the first nnn natural numbers:

Hⁿ = 1 + 1/2 + 1/3 + 1/4 ………..

What is the Fibonacci series and how is it generated?

The Fibonacci series is a sequence of numbers where each term is the sum of the two preceding ones, starting from 0 and 1. The first few terms of the Fibonacci series are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34.