__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

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