Creative thoughts-14


We know that square of a number n can be expressed as a sum of n number
of successive odd numbers from 1.The higher powers of a number also have
such a kind of properties.

Cube of a number n can be expressed as a sum of n numbers of
successive odd numbers which begins from n(n-1) +1,
i.e., (nxn)-n +1. It is interesting to note that the cubes of successive
 numbers show series with odd numbers. e.g.,

2x2x2 = 8 = 3+5= [(2x2)-1] + [(2x2)+1]
3x3x3 = 27 = 7+9+11= [(3x3)-2]+ (3x3) + [(3x3)+2]
4x4x4 = 64 = 13+15+17+19= [(4x4)-3]+
                                [(4x4) -1]+[(4x4)+1]+[(4x4)+3]
5x5x5 = 125 = 21+23+25+27+29= [(5x5)-4]+[(5x5)-2]+
                        (5x5)+[(5x5)+2]+[(5x5)+4]

The fourth power of a number n can be expressed as a sum of n
number of successive odd numbers which begins from (n-1)n(n+1) +1,
i.e., (nxnxn)-n +1,e.g.,

2x2x2x2 = 16 = 7 + 9 = [(2x2x2)-1]+[(2x2x2)+1]
3x3x3x3 = 81 = 25+27+29 = [(3x3x3)-2]+(3x3x3)+ [(3x3x3)+2]
4x4x4x4 = 256 = 61+63+65+67= [(4x4x4)-3]+
                                               (4x4x4)- 1]+[(4x4x4)+1]+[(4x4x4)+3]

As the fourth power of a number is also a square of a square number,
 it can be expressed as a sum of nxn odd numbers in natural series.

The fifth power of a number n can be expressed as a sum of n number of
successive odd numbers which begins from (nxnxnxn)-n+1,e.g.,

2x2x2x2x2 = 32=15+17
3x3x3x3x3= 243 = 79+8l+83
4x4x4x4x4= 1024= 253+255+257+259

Series of cubes

The sum of cubes of numbers in natural series is found to be square of the
sum of its roots. e.g.,

(1x1x1)+(2x2x2) = 1 + 8 = 9 =(1+2) x(1+2) = 3x3

(1x1x1)+(2x2x2)+(3x3x3) = 1+8+27 = 36 = (1+2+3)(1+2+3) = 6x6

(1x1x1)+(2x2x2)+(3x3x3)+(4x4x4)= 1+8+27+64 = 100 = 10x10

(1x1x1)+(2x2x2)+(3x3x3)+(4x4x4)+(5x5x5) = 225 = 15x15

We know the sum of numbers from 1 to n in natural series is n (n+1)/2
and its square is [nxn][(n+1)(n+1)]/4. From this facts, it can be shown
that for the sum of cubes of numbers in natural series, the sum of the
cube roots is invariably its factor. If the sum of cubes of numbers in
 natural series is divided by this factor, the dividend will be same as
 that of the factor.

Natural odd series is a series of odd numbers from 1 with a common
difference 2.Let us now find the sum of a natural odd series with a common
difference d.

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

In terms of the largest number N(H) = (1+nd) of the series, the sum becomes,

S = [N(H)+d-1][N(H)+1]/2d

If the odd series begins with an odd number ‘a’ and with a common difference d

S = a +(a+d) + (a+2d) + ………. (a+nd)

                = (n+1)a + d n(n+1)/2 = [(n+1)/2][2a+nd],
and in terms of the largest number N(H)= (a+nd)

S = [N(H)+d-a][a+N(H)]/2d