By using the special properties associated with the square numbers, one can establish the formula for the sum of successive squares. The square of a  number n can be expressed as a sum of natural numbers  from 1 to n in the ascending order and from (n-1) to 1 in the descending order.

1² = 1 =  1

2²= 4 =  1 + 2 + 1

3²= 9 =  1 + 2 + 3 + 2 + 1

4²= 16 =1 + 2 + 3 + 4 + 3 + 2 = 1

 The sum of all the four squares is 4(1) +3(2) +2(3) +1(4) +2(3) +3(1) = 7(1) +5(2) +3(3) +1(4)

In general the sum of all squares upto n in natural series is 1(n) + 3(n-1) + 5(n-2) + 7(n-3)…..(2n-1)(1). The terms in the total sum can be rearranged conveniently to derive the formula.

                                            1²                                                            1

                                            2²                                                    1 + 2  + 1

                                            3²                                     1 + 2 + 3 + 2 + 1

                                            4²                                     1 +2 + 3 + 4 + 3 + 2 + 1

 

There are one natural series from 1 to 4, 2 smaller series from 1 to 3, 2 smaller series from 1 to 2 and 2 (1). In general the sum of squares in natural series upto n

                      One natural series from 1 to n =    n(n+1)/2

                      Two natural series from 1 to n-1 = (n-1)n

                      Two natural series from 1 to (n-2) = (n-2)(n-1)

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

                      Two natural series upto 1                 = 1x2                                 

 

The total sum of all such broken series is  n(n+1)/2  + [(1x2) + (2x3) + (3x4) ….. (n-1)n]

 [(1x2) + (2x3) + (3x4) ….. (n-1)n] = (n-1)n(n+1)/3

 

   1²+ 2²+ 3²+ …….. n²  = n(n+1)/2  + (n-1)n(n+1)/3

                                      = n(n+1)/2 [1 + (2/3)(n-1)]

                                      = n(n+1)/2 [(2n+1)/3]

The very same relation can be derived with the help of yet another numerical property of squares.

 1² = 1 =  1

2²= 4 =    2 + 2

3²= 9 =    3 + 3 + 3

4²= 16 = 4 + 4 + 4 + 4

In general the sum of squares in natural series upto n

                      One natural series from 1 to n =    n(n+1)/2

                      One natural series from 2 to n =   n(n+1)/2 – (1x2)/2

                      One natural series from 3 to n =   n(n+1)/2 – (2x3)/2

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

                      One natural series with single term n = n(n+1)/2 – (n-1)n/2 

The total sum of all such broken series is  n²(n+1)/2 - (1/2)[(1x2) +(2x3) + (3x4) …. (n-1)n]

 since [(1x2) + (2x3) + (3x4) ….. (n-1)n] = (n-1)n(n+1)/3  

   1²+ 2²+ 3²+ …….. n²  = n²(n+1)/2 -(1/2)[(n-1)n(n+1)/3]

                                      = n(n+1)/2 [n – (n-1)/3]

                                      =[n(n+1)/2] [(2n+1)/3]

                                        

 Sum of series of successive squares

 S = 1² + 2² + 3² + 4² +……….   + n² , where n is the order of the number in the series which is equal to its numerical value

These squares are grouped into series of odd squares 1² + 3² + 5² +……….   + (2n-1)²

And series of even squares 2² + 4² + 6² +……….   + (2n)²

Where n is the order of the odd/even number in the series with numerical values  N_oH = (2n -1) and N_eH = 2n. It is noted that the sum of squares of all odd numbers from 1 to (2n-1) in natural series is 1/6 times the product of three successive numbers (2n-1)2n(2n+1)

 1² = (1x2x3)/6 = 1

1² +3² = (3x4x5)/6 = 10

1² +3² + 5² = (5x6x7)/6 =35 and so on.

1² + 3² + 5² +……….   + (2n-1)² = (2n-1)2n(2n+1)/6= (n/3)(4n² – 1)

Since N_oH = 2n -1 or 2n = (N_oH + 1) and 2n+1 = N_oH +2. In terms of N_oH ,the sum becomes Sodd² = N_oH (N_oH +1) (N_oH +2)/ 6

 

Like wise the sum of all even numbers from2 to 2n in the natural series is 1/6 times the product of three successive numbers 2n (2n+1) (2n+2). For example,

 

2²  = (2x3x4)/6  = 4

2² +4² = (4x5x 6)/6 = 20

2² +4² + 6² = (6x7x8)/6 = 56

In general 2² + 4² + 6² +……….   + (2n)²  = 2n (2n+1)(2n+2)/6 = (2/3) n (n+1)(2n+1)

Since N_eH = 2n or n = (N_eH /2) , (n+1) = (N_eH +2)/2 and 2n+ 1 = N_eH +1. In terms of N_eH, the sum becomes S_even² =(1/6) N_{eH (}N_eH+1) ₍N_eH +2)

The sum all squares in natural series is the sum of the sum of odd squares and even squares S² = S_odd² + S_even²

In terms of order the numbers (n/3) (4n² -1) + (2n/3)(n+1)(2n+1) = (n/3)(2n+1)(4n+1) . In terms of the numerical value of the number N_H = 2n

S = N_H (N_H +1)( 2N_H +1)/6

 

There is yet another way to derive the general expression for the sum of squares in natural series S = 1² + 2² + 3² + 4² + ………… This can be considered as an arithmetic series with different common difference i.e., S = 1 + (1+d₁) + (1+d₂) + (1+d₃) + ……. + (1+d_n-1) = n + [d₁ + d₂ + d₃ …….. d_n-1] or with same common difference between any two successive terms i.e., S =  1 + (1+d) + (1+2d) +  (1+3d) + ……. + [1+(n-1)d] = n + d[ 1+2+3…….(n-1)] = n + d[ n(n-1)/2] . Comparing these two expressions representing the same sum ,we have <d> = 2[d₁ + d₂ + d₃ …….. d_n-1] /n(n-1)

 

 By estimating the mean common difference, we can use the same formula used for arithmetic series with a common difference. The common difference may be different, but it varies gradually, which can be estimated.

 

1² + 2² = 5 = 1 +(1+d₁)  = 2 + d₁  or d₁ = 3; 1 + (1+d) = 2 + d or <d> = d₁ = 3

1² + 2² + 3²  = 14 = 3+ d₁ + d₂ which gives d₁ + d₂ = 11 and d₂ = 8

    3+ d₁ + d₂ = 3 +3d or d = 11/3 = 3 +(2/3)

1² + 2² + 3²  + 4² = 30 = 4 + d₁ + d₂  + d₃ = 4 + 6d ; d₁ + d₂  + d₃ = 26  and d₃ = 15. d = 13/3 = 3+(2/3) +(2/3) = 4 + (1/3)

1² + 2² + 3²  + 4² + 5² =55 = 5 + d₁ + d₂  + d₃ + d₄ = 5 +10d ; d₁ + d₂  + d₃ + d₄ = 50 and d₄ =24 , d = 4 + (1/3) + (2/3) = 5.

It is noted that the mean <d> is gradually increasing 3 + (2/3) (n-2)

If there are n terms S = n[ 1 + (n-1)<d>/2]

Substituting the mean value of <d>, we get,

S = n { 1+ [(n-1)/2][ 3 + 2/3(n-2)]} = n { 1 + [(n-1)/2](9+2n-4)/3 } = n [1 +(n-1/2)(2n+5/3)] = n[(2n² +3n +1)/6 = n(n+1)(2n+1)/6   

 

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].