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# 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 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

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: 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 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

…………………………………

…………………………………

………………………………..

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

Question 1. Find the sum of the n terms of the series whose nth terms is n2 + 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

Sn =(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: 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 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

…………………………………

…………………………………

………………………………..

(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

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

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