Special Series – Sequences and Series | Class 11 Maths

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 see how to get the formula for all these series.

Special Series 1: Sum of first n natural numbers

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 

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 

Special Series 2: Sum of squares of the first n natural numbers

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 

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

Special Series 3: Sum of cubes of the first n natural numbers

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 

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