User:Kiatgak

Sum of cubes in Archimedes way

First part, to decompose each cubic into a sum of squares, then compose them from different direction.

Let $S_n = \sum_{k=1}^n k^2 = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n$

so $n^3 = 3 S_n - \frac{3}{2}n^2 - \frac{1}{2}n$

apply it to $n^3 ... 1^3$

$n^3 = 3 [1^2 + 2^2 + ... + (n-1)^2 + n^2] - \frac{3}{2}n^2 - \frac{1}{2}n$

$(n-1)^3 = 3 [1^2 + 2^2 + ... + (n-1)^2 ] - \frac{3}{2}(n-1)^2 - \frac{1}{2}(n-1)$

...

$2^3 = 3 [1^2 + 2^2 ] - \frac{3}{2}2^2 - \frac{1}{2}2$

$1^3 = 3 [1^2 ] - \frac{3}{2}1^2 - \frac{1}{2}1$

To sum these n identities, let

$R_n = \sum_{k=1}^n k^3 = 3 [1^2*n + 2^2*(n-1) + ... + (n-1)^2*2 + n^2 *1] - \frac{3}{2}S_n - \frac{1}{2}T_n$

$R_n = 3 X_n - \frac{3}{2}S_n - \frac{1}{2}T_n$

where $T_n = \sum_{k=1}^n k = 1 + 2 + ... + n = \frac{1}{2}n(n+1)$

$X_n = 3 [1^2*n + 2^2*(n-1) + ... + (n-1)^2*2 + n^2 *1]$

Second part, make use of $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$; apply it to $(n+1)^3$:

$(n+1)^3 = [ 1 + n ]^3 = 1^3 + 3*1^2*n + 3*1*n^2 + n^3$

$(n+1)^3 = [ 2 + (n-1) ]^3 = 2^3 + 3*2^2*(n-1) + 3*2*(n-1)^2 + (n-1)^3$

...

$(n+1)^3 = [ n + 1 ]^3 = n^3 + 3*n^2*1 + 3*n*1^2 + 1^3$

To sum these n identities,

$n(n+1)^3 = R_n + 3 \left[1^2*n + 2^2*(n-1) + ... + (n-1)^2*2 + n^2 *1 \right] + 3 \left[ 1^2*n + 2^2*(n-1) + ... + (n-1)^2*2 + n^2 *1 \right] + R_n$

$n(n+1)^3 = 2R_n +2 * 3 X_n$

OK, easy part now, to merge the results of first part and second part, we can get

$\sum_{k=1}^n k^3 = \frac{1}{4}n^2(n+1)^2$

BTW, Archimedes would use $(n-1)^3$ at first part, and use $n^3$ at second part.

• Archimedes: What Did He Do Beside Cry Eureka