OPERATION OF SPECIAL SERIES

A series in mathematics is a crude description of the act of adding infinitely many numbers one after the other to a specified initial quantity. Series analysis is an important aspect of calculus and its generalization, mathematical analysis. Series are employed in almost all fields of mathematics, including the analysis of finite structures via generating functions. In addition to their widespread application in mathematics, infinite series are also frequently employed in physics, computer science, statistics, and finance.

In arithmetic, a sequence is a collection of items in which the order of the elements is important. The element or component in this case follows a precise pattern. In real life, we encounter sequences in a variety of contexts. House numbers in a row, pay in successive years (by a predetermined amount or by a certain percentage), page numbers in a book, and so on are examples of sequences.

 The sum of all the numbers in the given sequence is defined as a series. The sequences are both finite and endless. Similarly, the series might be finite or infinite. Consider the following sequence: 1, 3, 5, 7,… The sequence of these terms will therefore be 1 + 3 + 5 + 7 +… The series is referred to as a special series if it is unique in some respects. The three sorts of special series are as follows.

o   sum of first n natural numbers :- 1+ 2 + 3 +… + n

o   sum of squares of the first n natural numbers:- 12 + 22 + 32 +… + n2

o   sum of cubes of the first n natural numbers:- 13 + 23 + 33 +… +n3

SUM OF N NATURAL NUMBER:-

The sum of first n natural number can be calculated as:-

1+ 2 + 3 + 4 + …. + n = n (n + 1) / 2

Proof:- Sn = 1 + 2 + 3 + 4 +… + n

We can see that this is an Arithmetic Progression, with the initial term (a) equal to one and the common difference (d) equal to one, and there are n terms.

As a result, the sum of n terms is n/2 (2 x a + (n – 1) x d).

We will receive the values for this series if we put them together.

Sn = n/2(n – 1) x 1 + (n – 1) x 1)

Sn = (n + n – 1)/2(2 + n – 1)

 Sn = n(n + 1)/2

Hence the result of the first n natural number is n(n + 1)/2

Sum of the first n natural numbers’ squares:-

Sum of square of first “n” natural number can be calculated as:-

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

Proof :- Let us take  Sn = 12 + 22 + 32 +… + n2   —eq 1

We know that k3 – (k – 1)3 = 3k2 – 3k + 1  — eq 2

Also, (a – b)3 = a3 – b3 – 3a2b + 3ab2

So, k3 – (k – 1)3

= k3 – k3 +1 + 3k2 – 3k

= 3k2 – 3k +1

Now Putt k = 1, 2…, n successively in eq 2, we get

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

Now on 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 be written as,

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

The Sum of the natural numbers’ cubes from 1 to n:-

Putting k = 1, 2…, n,  one by on successively in eq 2, we obtain

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

And also, (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 the eq 2 , we get,

(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

Now on Adding both the 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 be written as,

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.

CONCLUSION:-

A series is the cumulative sum of a particular sequence of terms in mathematics. , a sequence is a collection of items (usually integers) in which the order of the components is important. The components in this case follow a precise pattern. In real life, we encounter sequences in a variety of contexts. House numbers in a row, pay in successive years (by a predetermined amount or by a certain percentage), page numbers in a book, and so on are examples of sequences.