My Contribution to the Mathematical Arts

[updated 4/10/25 to use MathJax]
[updated 4/10/25 - Grok claimed that formula (9) was wrong, which I subsequently verified]
[updated 4/11/25 - Formula (9) is, in fact, correct.]

Many, many years ago, sometime in high school, I learned that the formula for computing

(1) $1 + 2 + 3 + \dots + n$

is

(2) ${\Large \frac{n(n + 1)}{2}}$

I'm pretty sure we were shown the geometrical derivation of this formula. In college, in late 1975 or early '76, in a Math Lab one problem asked to guess the formula for

(3) $1^2 + 2^2 + 3^2 + \dots + n^2$

In my handwritten lab notebook I used induction to show that the solution to (3) is:

(4) ${\Large \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}}$

Having solved this specific case, I wanted to see if there was a general solution for any positive integer n and any positive integer exponent. Based on equations (2) and (4), I conjectured that the general solution would be a polynomial of this form:

(5) $\displaystyle \sum_{i=1}^{n} i^k= a_{k+1}n^{k+1} + a_{k}n^{k} + a_{k-1}n^{k-1} + \dots + a_{2}n^{2} + a_{1}n$

The
derivation used induction on the general formula and found that the coefficients to the solution are:

(6) $a_i = \frac{\binom{k}{k-i+1} - \displaystyle \sum_{j=1}^{k-i+1} a_{i+j} \binom{i+j}{j+1}}{i} \quad 1 \leq i \leq k+1$

Computed coefficients for $0 \leq k \leq 15$ are
here.

Perhaps of interest are these properties of the coefficients:

(7) $a_{k+1} = {\Large \frac{1}{k+1}}$

(8) $a_{k} = {\Large \frac{1}{2}}$    when $k > 0$

(9) $a_{k-1} = {\Large \frac{k}{12}}$    when $k > 1$

(10) $a_{k-2} = 0$     when $k > 2$

(11) $\displaystyle \sum_{i=1}^{k+1} a_i = 1$
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