## SIVAHARI THEOREM ON ODD AND EVEN NUMBERS

Posted by haridas | Uncategorized | Posted on March 11th, 2013

Abstract

Squares of integers can be expressed as sum of consecutive odd numbers.  But a general case of all powers,  after a thorough search , could not be found. Here is an attempt in that lines. Sum of consecutive odd numbers  as powers of integers and sum of consecutive even numbers also as powers of integers with formulae and simple proof.

Sivahari Theorem on odd integers.

1.1 All powers of  positive  integers  can  be  expressed  as  sum of consecutive   odd  integers,  where  the  number  of  terms  will be equal to  the number  itself.

Ex:

1.2

1+3+5+7+…………+ (2n-1)       = n2

1.3

1                                               =13

3+5                                          = 23

7+9+11                                    =33

13+15+17+19                         =43

21+23+25+27+29                = 53

1.4

1                                              = 14

7+9                                          = 24

25+27+29                              = 34

61+63+65+67                       = 44

121+123+125+127+129      = 54

1.5

1                                               = 15

15+17                                      = 25

79+81+83                              =35

253+255+257+259              = 45

621+623+625+627+629    =55

1.6

In   general  when  n  and  x  are positive  integers

[n x-1-(n-1)] +   [n x-1-(n-3)] + [n x-1-(n-5)] +……………+  [n x-1+(n-3)] +  [n x-1+(n-1)] = nx

This formula will  generate  the  required  power and  its sequence.

Proof

[n x-1-(n-1)]+ [n x-1-(n-3)]+ [n x-1-(n-5)]+……………+[n x-1+(n-3)] + [n x-1+(n-1)]

= n( n x-1 )  – (n-1) – (n-3) – (n-5) +……….+ (n-5)  + (n-3) + (n-1)

= nx

This can also  be  proved  as  an  Arithmetic Progression  with common  difference  2.

Sn = [(first term + last term) n] / 2

= {[n x-1 – (n-1) + n x-1 + (n – 1)] n} / 2

=   n.n x-1

=   nx

Sivahari Theorem on even integers

2.1

For integers  n>1 and  x = 2  all  squares  of  numbers  can be expressed as the sum of consecutive   even  integers  minus that number, where  the number of  terms  will be  equal  to  the number  itself.

2.2

2 + 4 + 6 + 8 +…………+ 2n = n (n + 1) = n2 + n

3.1

For integers  n>1 , x>2   all powers of numbers  can be expressed  as sum of consecutive even integers  plus that number, where  the  number  of terms  will be  equal  to  the number  itself.

3.2

2 +4                                           = 23 – 2

6 + 8 +10                                   =  33- 3

12 + 14 +16 +18                        = 43 – 4

20 + 22 +24 + 26 + 28           = 53 – 5

3.3

6 + 8                                             =  24 – 2

24 + 26 +28                                =  34 – 3

60 + 62 + 64 + 66                     =  44 – 4

120 + 122 + 124 + 126 +128   = 54-  5

3.4

14 + 16                                           = 25-2

78 + 80 +82                                 = 35- 3

252 + 254 + 256 + 258              = 45 – 4

620 + 622 + 624 + 626 +628   = 55 – 5

3.5

In general for integers n > 1 and  x > 2  we have ,

(n x – 1 - n) + (n x – 1 - n + 2) + (n x – 1- n + 4)+…………..+(n x – 1 + n – 2) = nx -n

This formula gives the required power and its sequence .

3.6 Proof

Since this sequence represents an  AP  with common difference 2

Sn =  [( first term + last term ) n] / 2

= [ ( n x-1 – n     + n x-1 + n - 2 ) n] / 2

=    [ ( 2 n x-1 – 2 ) n] / 2

=    n.n x-1 –n

=   nx – n

Hence the general theorem for odd and even  integers .

S. Haridasan

(near Vyapara Bhavan)

Parippally P.O.

Kollam, 691574

Kerala State

INDIA.

Phone: 91 9447015945

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