Carl Friedrich Gauss, German  mathematician and physicist when he was doing his school education ,once his teacher gave an arithmetic problem , the sum of all numbers from 1 to 100  as class work to all students .The teacher thought the students will do the sum one by one in the conventional way and they will take some time to finish. But amazingly Gauss worked out the problem within no time. When he showed the correct answer to his teacher   the teacher asked him “how you did?” Gauss told in the given arithmetic progression, the first and last number i.e., 1^st and 100^th terms ,2^nd and 99^th terms, 3^rd and 98^th term are summed up to 101 and likewise all the remaining numbers can also be paired up to give a sum 101. There are 50 such pairs. Hence the total sum must be equal to 50 times 101 or 5050. In general, S_n = 1 + 2 + 3 + 4………. n = ₁ⁿ ∑n   = n (n+1)/2. This was the beginning of finite arithmetic series, and introduced as one of the chapters in school mathematics.

The sum of the numbers in natural series (Gauss method)

     The sum of the numbers in natural series from 1 to n is S_n = 1+2+3+4+..... (n-1) + n = ₁ⁿ∑n. It is seen that the sum the first and last terms is (n+1). And likewise the sum of the second and last but one term is (n+1). If n is even , i.e., the number of terms in the summation is even, all the numbers can be paired up and there will be (n/2) number of such pairs with (n+1) as a sum of each pair. The sum of the series then becomes S_n = (n/2)(n+1).If n is odd. There will be (n-1)/2 pairs whose individual sum is (n+1) and the middle term (n+1)/2 is unpaired. In such case the sum of the series becomes (n-1)/2 [n+1] + (n+1)/2 = (n+1)/2 [n-1+1] = n(n+1)/2.

      This can be proved by summing up two series of S_n with terms in ascending and descending orders

                            S_n =   1 +     2   +     3   + 4 + ………………… + (n-1) +    n

                            S_n =   n  + (n-1) + (n-2) + …………………     +    2     +    1

                           -----------------------------------------------------------------------------

                           2S_n = (n+1) +(n+1) +(n+1) +…………………+ (n+1) + (n+1)

                                  = n (n+1)

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

Since the number of terms n represents N, the number itself, S_n = (N/2)(N+1)  

 

Bound natural series:

     The sum of the numbers in natural series from n_L to n_H where n_L is lower or beginning   number and n_H is higher or ending number

S =  n_{L + (}n_{L+1) + (}n_{L + 2) + ………..}  n_H

If S_H and S_L-1 denote the sum if sum of numbers in natural series from 1 to n_H and from 1 to n_L- 1 respectively we have

 S_H = n_H (n_H +1)/2

S_L-1 = (n_L- 1) n_L/2

S = S_H- S_L-1 = [n_H (n_H +1)/2] – [(n_L- 1) n_L/2]

On simplification we get S = (n_H + n_L) ( n_H - n_L+ 1) /2

 

   This can also be proved by yet another method. In S = n_{L + (}n_{L+1) + (}n_{L + 2) + ………..}  n_H there are (n_H - n_L+ 1) terms in the series, where the sum of the first term (n_L) and the last term (n_H )  is (n_H  + n_L). If (n_H - n_L+ 1) is even there will be (n_H - n_L+ 1)/2 pairs whose individual sum is (n_L + n_H). Hence the sum is the product of number of pairs and its individual sum.

S = (n_H - n_L+ 1) (n_H + n_L)/2. If there are odd number of terms there will be n_H - n_L number of terms in  which the middle number (n_H + n_L)/2  is unpaired .Leaving this unpaired term,  the number +   of paired terms are  (n_H – n_L)/2   the sum of all paired members is  (n_H – n_L)/2 x (n_H + n_L) , while the single unpaired middle term is (n_L + n_H)/2 .Total sum becomes (n_H – n_L)/2 x (n_H + n_L) + (n_L + n_H)/2 = [(n_L + n_H)/2][ n_H – n_L+ 1]

 

The sum of even numbers in natural series:

 

    The natural series of even numbers is [2 + 4 +6 +8 +10+……. 2n = 2[1+2+3+4……….n] where n denotes the number of terms in the series. The sum of n even numbers in natural series then becomes 2 [n(n+1)/2]  = n(n+1). If N is the last number N = 2n . Then the sum of n number of even numbers in natural series in terms of N, the final number (N_/2)[(N + 2)/2] =N (N +2)/4.

     By using Gauss method   S_even = 2    +   4    + 6 +     8 + ….   2n    in ascending order

                                                 S_even= 2n + (2n-2) + (2n-4) + ….   2       in descending order

                                                -----------------------------------------------------------------

                                                2S_even = 2(n+1) + 2(n+1) + ……     2(n+1)

                                                         = n x 2(n+1)               

                                       Or S_even = n(n+1)

 

The sum of odd numbers in natural series:

 

       The natural series of odd numbers is 1 +3 + 5 +7 ……. (2n-1)

We note that 1 +3 = 4 = 2²

                                  1 + 3 + 5 = 9 = 3²

                                  1 + 3 + 5 + 7 =16 = 4²

In 1+ 3 + 5 + 7 +….. (2n-1) there are n terms and hence 1 + 3 + 5 + 7…..(2n-1) = n². The sum n odd numbers from 1 to (2n-1) is a square of a number which is equal to the mean of the first and last odd numbers i.e., [(1 + 2n-1)/2]² = n²

This can be proved logically by converting the given odd series into even series. If we remove 1 from each term, all the terms will become even.

      S_odd = 1 + 3 + 5 + 7 + ……  (2n-1)

             = 1 + (1+2) + (1+2x2) + (1+ 3x2) …….. [1+ 2(n-1)]

Since there are n number of terms in the series, the 1 from each terms will contribute n to its   sum S_odd. The remaining terms form an even series 2 + 4 +6 ……. (2n-2) = 2[1+2 +3 …….(n-1)]. The terms inside the bracket are in natural series with (n-1) terms. Hence its sum is 2 [n(n-1)/2] = n(n-1). Adding with n, the sum of n number of l removed from each term, we get  Sn = n² , since (2n-1) = N then S_n  in terms of N _, S_n = [(N +1)/2]²

 

      The sum of all numbers in the given natural series is 1+2+3+4 …….N = N(N+1)/2. If N is an odd number then N = (2n-1). The given series is separated into two different series with odd and even numbers only upto N. Then,

     1+2+3+4+…… N = [1+3+5+…. (2n-1)]+ [2+4+6+ (2n-2)]

                n(2n-1)    =     n² +  [n(n-1)]

 In terms of N ,  N(N+1)/2 =  [(N+1)/2]² + 2{1+2+3….. (N-1)/2]

                                             = [(N+1)/2]² + (N² -1)/4

The sum of odd numbers in natural series is the difference between the sum of all numbers   and the sum of all even numbers in the given natural series.

 

        [1+3+5+…. N] =   N(N+1)/2 – (N²- 1)/4 =  [(N+1)/2]²

            [1+3+5+…. (2n-1)]= n(2n-1) – n(n-1) =  n²

 

If N is an even number in the given natural series, then N = 2n and

     1+2+3+4+…… N = [1+3+5+…. (2n-1)]+ [2+4 +6+ …..2n]

                            n(2n+1)  =  n²+ n(n+1)

In terms of N , N(N+1)/2 = (N/2)²  + N(N+2)/4

The difference between the sum of all numbers and the sum of all even numbers in the given series gives the sum of odd numbers in natural series.

       [1+3+5+…. (2n-1)]= n(2n+1) – n(n+1)  =  n²

           [1+3+5……(N-2)/2] = N(N+1)/2 – N(N+2)/4  =  (N/2)²

 

Conversion method:   

    By converting the even series into odd or the odd series into even, one can compute sum of the given series.    Let us consider the general form of even series

                             2 + 4 + 6 + ……… 2n

By separating 1 from each number the above series is changed into odd  

 

2 + 4 + 6 + ……… 2n = (1+1) + (1+3) + (1+5) ……..   [1+(2n-1)]

                                    = n + [ 1+ 3 + 5 +….. (2n-1)]

                                    = n + n²

                                                    = n(n+1)

In terms of N  , 2 + 4 + 6 + ……… N = N(N+2)/4 

The same result is obtained if we add 1 to each number in the given series

 2 + 4 + 6 + ……… 2n = (3-1) + (5-1) + (7-1) ……..   [(2n+1)-1 ]

 

                                      = [3+5+7+…….. (2n+1)] – n

                                      = [1 +3+5+7……..(2n+1)] –(n+1)

                                     = (n+1)² – (n+1)

                                      = n (n+1)

Likewise an odd series can be converted into an even series either by adding 1 to each term or  subtracting 1 from each term . Considering the general form of an odd series

                                  1 + 3+ 5 +     (2n-1)

By adding 1 to each term, it is converted into even series

1 + 3+ 5 +     (2n-1) = (2-1) + (4-1) + (6-1)…….   (2n-1)

                                 = [2 + 4 + 6 +…. 2n] – n

                                  = n(n+1) – n  =   n²   

 The general formula for the sum of successive odd squares from 1 to (2n-1) . It is supposed that the series end with  square of an even number.

S_{n  =}    1²  + 2² + 3² +4² + 5 ²   + 6²  + 7² + 8² +………   (n-1)² +(n)² =  n (n+1) (2n+1)/6

 The last number L =  n ; S_n = L(L+1)(2L+1)/6

The given finite series is split into two series one with odd squares and another with even squares.

1²  + 2² + 3² +4² + 5 ²   + 6²  + 7² + 8² +………   (n-1)² +(n)² =  [1²  + 3² + 5 ² + 7² +……   (2m-1)² ]+ [ 2² + 4² + 6²  + 8² +………    +(2m)²]

[1²  + 3² + 5 ² + 7² +……   (2m-1)² ] = [1²  + 2² + 3² +4² + 5 ²   + 6²  + 7² + 8² +………   (n-1)² +(n)²] – [2² + 4² + 6²  + 8² +………    +(2m)²]

          = [n(n+1)(2n+1)/6] -  4 (n/2)[(n+2)/2] (n+1)/6   =  n (n² – 1)/6  

In terms of last number L  1²  + 3² + 5 ² + 7² +……   (2m-1)² ] = L(L+1)(2L+1)/6 – L(L+1)(L+2)/6 =  L(L²- 1)/6

 

∑(2n)³  = k_n ∑(2n) என்று கொண்டு ∑(2n)³க்கான சூத்திரத்தை வருவிக்கலாம் .

  n          ∑2n           ∑(2n)³             k_n                    Δ k_n                                                                         ……………………………………………………………………

                       1                 2                8                4              ……

                       2                 6              72              12                8

                      3                12            288              24              12

                      4                20            800              40              16          

                      5                30           1800             60              20            

               ……………………………………………………………………………..

  k_n = 4 + 8 + 12 + …….. 4n = 4[1+2+3+……n] = 2n(n+1)

∑(2n)³  = k_n ∑(2n)  = 2n (n+1) ∑(2n) = 2n (n+1) [n(n+1)] =  2n² (n+1)² . n என்பது தொடரில் உள்ள எண்களின் எண்ணிக்கை. 2n = N என்பதால் ∑(N)³   = (N²/8)[N+2].

 

1 முதல் n வரையுள்ள  இயல் தொடரின் கூட்டுத்தொகை

1+ 2 = 3 =   (1+1/2) + (2-1/2) = 2 x 3/2 = 3
கூட்டப்படும் இரு எண்களின் சராசரி மதிப்பு (1 +2)/2 = 3/2                     

1+ 2+ 3+……. n  =    ∑ n 

கூட்டப்படும்  எண்களின் சராசரி மதிப்பு  ∑ n / n = (n+1)/2

n இரட்டை மதிப்புடையதாக இருந்தால் கூட்டலில் உள்ள ஒரு எண்ணின் சராசரி மதிப் பு  (n +1 )/2  . எனவே n எண்களின் கூட்டுத் தொக n (n+1)/2 . அல்லது கூட்டலில் (n+1)   கூட்டுத் தொகைகொண்ட (n /2) இணைகள்  இருப்பதால் கூட்டுத் தொகை n(n+1)/2  ஆகும் n ஒற்றை எனில், ஒன்றின்  மதிப்பு (n + 1 ) கொண்ட (n -1 )/2 எண் இணைகள் இருக்கும். இணையற்ற மைய எண்ணின் மதிப்பு (n +1) /2. இவற்றின் கூட்டுத் தொகை  n(n+1)/2.

                       இத்தொடர்பையே அடிப்படையாகக் கொண்டு  1 முதல் n வரையுள்ள எண்களின் இருமடி மற்றும் மும்மடிகளின் கூட்டுத்தொகைக்கான சூத்திரத்தை நிறுவமுடியும். 1 முதல் n வரையுள்ள எண்களின் இருமடிகளின் கூட்டுத்தொகை , 1 முதல் n வரையுள்ள எண்களின்  கூட்டுத்தொகைக்கு நேர் விகிதத் தொடர்பில் இருப்பதாகக் கொள்ளலாம்.

                                    ∑ n² = k_n ∑ n

n ன் வெவ்வேறு மதிப்புக்களுக்கு k_n  மதிப்பை அறிந்து n  க்கு ஏற்ப மாறுபடும் k_n  மதிப்பை நிறுவமுடியும்

                n          ∑ n                 ∑ n²                   k_n                    Δ k_n

               …………………………………………………………………..

               1               1                      1                  1              ……

               2               3                      5                1 2/3            2/3

               3               6                     14               2 1/3           2/3

              4              10                     30               3                2/3

              5              15                    55               3 2/3             2/3

            ……………………………………………………………………….

k_n  = 1 + (n-1) (2/3) = (2n+1)/3

∑ n² = k_n ∑ n = [(2n+1)/3][ n (n+1)/2] = n(n+1) (2n+1)/6

இது போல    ∑ n³ = k_n ∑ n

n ன் வெவ்வேறு மதிப்புக்களுக்கு k_n  மதிப்பை அறிந்து n  க்கு ஏற்ப மாறுபடும் k_n  மதிப்பை நிறுவமுடியும்

                n          ∑ n                 ∑ n³                   k_n                    Δ k_n

               …………………………………………………………………..

               1               1                      1                 1              ……

               2               3                      9                 3                 2

               3               6                    36                 6                3

              4              10                   100              10                4

              5              15                   225              15                   5

            ……………………………………………………………………….

k_n   = ∑ n

∑ n³ = k_n ∑ n = ∑ n X ∑ n = [∑ n]² = n²(n+1)²/4

    இந்த வழிமுறையைப் பின்பற்றி உயர் மடிகளால் ஆன இயலெண் தொடருக்கான சூத்திரத்தை ஏற்படுத்திக்கொள்ள முடியும்.

∑ n⁴ = k_n ∑ n²    என்று கொள்வோம்.

n          ∑ n²               ∑ n⁴             k_n                    Δ k_n                                                                         …………………………………………………………………………..

                    1               1                      1            1              ……

                   2               5                    17         17/5          12/5

                  3              14                    98           7             18/5

                 4               30                   354       59/5           24/5          

                 5               55                   979       89/5           30/5         

               ……………………………………………………………………………..

k_n  = 1 + (6/5) [2+3++4+5…….n] = 1 +(6/5) [n (n+1)/2  - 1]= 1 +(3/5)[n² + n – 2]

   = (3n² +3n – 1)/5

∑ n⁴ = k_n ∑ n² = [(3n² +3n – 1)/5] [n(n+1)(2n+1)/6]

∑ n⁵ = k_n ∑ n³    என்று கொள்வோம்.

n          ∑ n³               ∑ n⁵             k_n                    Δ k_n                                                                         …………………………………………………………………………..

                    1               1                      1            1              ……

                   2                9                    33         11/3          8/3

                  3               36                  276         23/3          12/3

                 4              100                 1300        13              16/3          

                 5              225                  4425       59/3           20/3         

               ……………………………………………………………………………..

k_n  = 1 + (4/3) [2+3++4+5…….n] = 1 +(4/3) [n (n+1)/2  - 1]= 1 +(2/3)[n² + n – 2]

   = (2n² +2n – 1)/3

∑ n⁵ = k_n ∑ n³ = [(2n² +2n – 1)/3] [n(n+1)/2]²

∑ n⁶ = k_n ∑ n²    என்று கொள்வோம்.

n          ∑ n²               ∑ n⁶             k_n                    Δ k_n                                                                         …………………………………………………………………………..

                    1               1                      1            1              ……

                   2                5                    65          13             12 = 84/7

                  3                14                  794         397/7       306/7

                  4                30                 4890        163           744/7          

                  5                55                 20515       373         1470/7           

               ……………………………………………………………………………..

k_n  = 1 + (6/7) [14+51+124+245+…….]  = (3n⁴ +6n³ –3n + 1)/7

∑ n⁶ = k_n ∑ n² = [(3n⁴ +6n³ –3n + 1)/7][n(n+1)(2n+1)/6]