Sum of Consecutive Cubes
Calculate the sum of first n cubes or the sum of consecutive cubic numbers from n_{1}^{3 } to n_{2}^{3 }. A cube number (or a cube) is a number you can write as a product of three equal factors of natural numbers. The sum of consecutive cubic numbers from n_{1}^{3 } to n_{2}^{3 } is equal to:
n_{1}^{3 } + (n_{1} + 1)^{3} + ... + n_{2}^{3}
for example we input these value 5 it means we wants the sum of 1^{3},2^{3},3^{3},4^{3}, and 5^{3}
= 1x1x1 + 2x2x2 + 3x3x3 + 4x4x4 + 5x5x5
= 1 + 8 + 27 + 64 + 125
= 225
now calculate the same example using following formula:
Sum of consecutive squares = (n^{2}(n + 1)^{2}/4 )
= 5x5 (5 + 1)(5 + 1)/4
= 25 (6x6)/4
= 25x36/4
= 25x9
= 225