Sum of successive square numbers

S = 1² + 2² + 3² + 4² +……….   + n² , where n is the order of the number in the series which is equal to its 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 values  N_o – (2n -1) and N_e= 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)

In terms of the odd numbers  N_oH = 2n -1, 2n = (N_oH + 1) and 2n+1 = N_oH +2. In terms of N_oH ,the sum becomes So² = 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)

In terms of the even numbers  N_eH = 2n , 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_e² =(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 sum of even squares S² = S_o² + S_e²

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 actual 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 . 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) = 1 + (1+d) or d= d₁ = 3

1² + 2² + 3²  = 14 = 3+ d₁ + d₂ = 3 + 3d; d₂ + d₂ = 11 and d₂ = 8 and 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  1 +(1+d) + (1+2d) + (1+3d) …….. [1 +(n-1)d]

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