Natural series of even numbers
It is interesting to find the sum of even or odd numbers only in the given natural
series. The sum of the even numbers from 2 to 2n containing n even numbers
is given by
S(even) = 2+4+6+8+ …….. 2n = 2(1+2+3+4+……n)
The terms inside the bracket are in natural series up to n. Hence its sum becomes
S(even) = 2 x n(n+1)/2 = n(n+1),where n is the number of even numbers added up
which is equal to half of the last even number in the given even series. Thus the sum,
in terms of the last even number N of the series is
S(even) = (2n) x(2n +2)/4 = N(N+2)/4
Natural series of odd numbers
The sum of the odd numbers from 1 to N (=2n-1) in the natural series is given by
S(odd) = 1+3+5+7+9+….. (2n-1),where n = (N+1)/2
But S(odd) = S(natural ) – S (even),the difference between the total sum of the natural
series from 1 to (2n-1) and the series of all even numbers from 2 to (2n-2).
The sum of the natural series from 1 to (2n-1) is n(2n-1) and in terms of the last
number of the series ,it is N(N+1)/2.Similarly ,the sum of all even numbers from 2 to
2(n-1) is n(n-1) and in terms of last number N, it is [(NxN)-1]/4. Hence the sum of all
odd numbers from 1 to (2n-1) becomes,
S(odd) = n(2n-1)- n(n-1) = nxn
Thus the sum of natural series of n odd numbers from 1 is always a square number nxn.
In terms of N ,the last odd number of the series
S(odd) = [N(N+1)/2] – [(NxN)-1]/4 = (1/4)[(N+1)(N+1)]
If the natural series has a set of numbers from N(L),the smallest number to N(H),
the greatest number, its sum is given by
S = N(H)[N(H)+1]/2 – N(L)[N(l)+1]/2
= [N(H)-N(L)][N(H)+N(L)+1]/2
In the case of both natural even series and natural odd series , it is
S= [N(H)-N(L)][N(H)+N(L)+2]/4
Series of square numbers
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 and (2n+1).e.g.,
1x1 = 1x2x3/6 = 1
1x1 + 3x3 = 3x4x5/6 = 10
1x1+3x3+5x5 = 5x6x7/6 = 35
1x1+3x3+5x5+7x7 = 7x8x9/6 = 84
for the product of any three successive numbers in natural series ,6 will
invariably be a factor. Hence with the help of Algebra, it can be shown as
S(odd square) = 1x1 +3x3+5x5 +……. (2n-1)x(2n-1)
= [(2n-1)2n(2n+1)]/6 = n (4nxn – 1)/3
In terms of the last number of the given natural series of square numbers, it becomes,
S(odd square) = N(N+1)(N+2)/6
Like this the sum of squares of all even numbers from 2 to 2n(=N) in the
natural series is 1/6 times the product of three successive numbers 2n,2n+1,2n+2, e.g.,
2x2 = 2x3x4/6 = 4
2x2 +4x4 = 4xx6/6 = 20
2x2 +4x4+6x6 = 6x7x8/6 = 56
and in general
S(even squares) = 2x2 + 4x4 + 6x6 + …… 2nx2n
= 2n(2n+1)(2n+2)/6 = (2/3) n(n+1)(2n+1)
The sum of squares of all numbers from 1 to N in the natural series
can also be expressed in a similar fashion.
S(natural squares) = S(odd squares) + S(even squares)
= N(N+1)(N+2)/6 + (N-1)N(N+1)/6
= N(N+1)(2N+1)/6
The sum is (1/6) times the product of two successive numbers N and N+1
which is multiplied with its sum (2N+1)
1x1+2x2+3x3+4x4 +…… NxN = N(N+1)(2N+1)/6
