Frequently Asked Questions
Is there a formula to add a sequence of cubes?
By examining the first five sums a remarkable discovery is suggested:
1^{3} = 1
1^{3}+2^{3}=9
1^{3}+2^{3}+3^{3}=36
1^{3}+2^{3}+3^{3}+4^{3}=100
1^{3}+2^{3}+3^{3}+4^{3}+5^{3}=225
It seems that the sum is always square, but what is even more remarkable is that the sum of the first n cubes, 1^{3}+2^{3}+...+n^{3} = (n(n+1)/2)^{2}, which is the square of the n^{th} triangle number.
For example, 1^{3}+2^{3}+...+10^{3}=(10×11/2)^{2}=55^{2} = 3025.
Using a similar method used to prove that formula for the Sum of Squares, we shall prove this result deductively; it is hoped that it will offer some insight into how further the series of powers may be found.
Proof

r^{4}–(r–1)^{4}  =  n^{4}–(n–1)^{4} + (n–1)^{4}–(n–2)^{4} + ... + 3^{4}–2^{4} + 2^{4}–1^{4} + 1^{4}–0^{4}  
=  n^{4} 
But r^{4}–(r–1)^{4} = r^{4} – (r^{4}–4r^{3}+6r^{2}–4r+1) = 4r^{3}–6r^{2}+4r–1.
∴ ∑ 4r^{3}–6r^{2}+4r–1  =  4∑ r^{3} – 6∑ r^{2} + 4∑ r – ∑1 
=  4∑ r^{3} – 6n(n+1)(2n+1)/6 + 4n(n+1)/2 – n  
=  4∑ r^{3} – n(n+1)(2n+1) + 2n(n+1) – n  
=  n^{4}. 
∴ 4∑ r^{3}  =  n^{4} + n(n+1)(2n+1) – 2n(n+1) + n 
=  n(n^{3} + (n+1)(2n+1) – 2(n+1) + 1  
=  n(n^{3} + 2n^{2}+3n+1 – 2n–2 + 1)  
=  n(n^{3}+2n^{2}+n)  
=  n^{2}(n^{2}+2n+1)  
=  n^{2}(n+1)^{2}  
∴ ∑ r^{3}  =  n^{2}(n+1)^{2}/4 
=  (n(n+1)/2)^{2} 
In other words, the sum of the first n cubes is the square of the sum of the first n natural numbers.