RSS
Frequently Asked Questions
Is there a formula to add a sequence of cubes?
By examining the first five sums a remarkable discovery is suggested:
13 = 1
13+23=9
13+23+33=36
13+23+33+43=100
13+23+33+43+53=225
It seems that the sum is always square, but what is even more remarkable is that the sum of the first n cubes, 13+23+...+n3 = (n(n+1)/2)2, which is the square of the nth triangle number.
For example, 13+23+...+103=(10×11/2)2=552 = 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
|
r4–(r–1)4 | = | n4–(n–1)4 + (n–1)4–(n–2)4 + ... + 34–24 + 24–14 + 14–04 | |||
| = | n4 |
But r4–(r–1)4 = r4 – (r4–4r3+6r2–4r+1) = 4r3–6r2+4r–1.
| ∴ ∑ 4r3–6r2+4r–1 | = | 4∑ r3 – 6∑ r2 + 4∑ r – ∑1 |
| = | 4∑ r3 – 6n(n+1)(2n+1)/6 + 4n(n+1)/2 – n | |
| = | 4∑ r3 – n(n+1)(2n+1) + 2n(n+1) – n | |
| = | n4. |
| ∴ 4∑ r3 | = | n4 + n(n+1)(2n+1) – 2n(n+1) + n |
| = | n(n3 + (n+1)(2n+1) – 2(n+1) + 1 | |
| = | n(n3 + 2n2+3n+1 – 2n–2 + 1) | |
| = | n(n3+2n2+n) | |
| = | n2(n2+2n+1) | |
| = | n2(n+1)2 | |
| ∴ ∑ r3 | = | n2(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.