Q13/002) Prove that arranging odd positive integers in the following way: 1 3 + 5 7 + 9 + 11 13 + 15 + 17 + 19 21 + 23 + 25 + 27 + 29 ......................................… produces the sequence of the cubes of the integer more precisely the sum of each line is the cube of the quantity of the numbers

The nth line has got n numbers.

now the 1st number
1,3 ,7 ,13, 21 ..

1st order difference 2,4,6, 8

2nd order difference = 2,2,2,

as it is constant we have this equation is quadratic in n

so an^2 + bn + c
putting n =1 ,2 and 3 for 1st 3 terms

n =1 => 1 = a + b+ c ..1
n= 2 => 3 = 4a + 2b + c ..2
n =3 => 7 = 9a + 3b + c ...3

from (1) and (2) 3a + b = 2
from (2) and (3) 5a + b = 4

so a =1 and b= -1 and hence c = 1

so term =n^2 - n + 1

now in the nth row 1st term = n^2 - n + 1 and common difference = 2 and n terms

so last term = 2n^2 - n + 1 + 2(n-1) = n^2 +n - 1

so the average = 1/2(n^2 - n + 1 + n^2 + n -1) = n^2
as there are n terms sum = n^3