Jakob Bernoulli was born in Basel, Switzerland, to a large family that would become well known for its mathematical talents. He and his younger brother, Johann Bernoulli (1667-1748), the best known mathematicians in the family, had a relationship that was at times collaborative and at times contentious. Both brothers carried on Leibniz’s work in calculus and supported Leibniz in his priority dispute with Newton.

Bernoulli’s book on probability, *Ars Conjectandi,* was published posthumously in 1713. In it, Bernoulli derived symbolic formulas for the sums of positive integer powers using the method conjectured above for Fermat, then noted a pattern that would make computation of these formulas much simpler (Bernoulli, pp. 95-98, or Smith, pp. 85-90). Bernoulli first showed how to find the sum of the first *n* positive integers, obtaining $$\int n = {1 \over 2}\,nn + {1 \over 2}\,n$$

(Smith, p. 87). (Note that he used an integral sign, where we use a "sum" sign.) He then derived the sums of the first *n* squares and cubes, obtaining $$\int nn = {1\over 3}\,n^3 + {1\over 2} \,nn + {1 \over 6}\,n \quad {\rm and} \quad \int n^3 = {1 \over 4}\,n^4 + {1 \over 2}\,n^3 + {1 \over 4}\,nn$$

(Smith, pp. 87-89). To obtain the sum of the first *n* cubes, Bernoulli noted first that the general term in the fourth column of his “table of combinations” (see Fig. 11) was $${{n - 1.\,n - 2.\,n - 3} \over {1.2.3}} = {{n^3 - 6nn + 11n - 6} \over 6}.$$

1 | 0 | 0 | 0 | 0 | 0 | 0 | ||||||

1 | 1 | 0 | 0 | 0 | 0 | 0 | ||||||

1 | 2 | 1 | 0 | 0 | 0 | 0 | ||||||

1 | 3 | 3 | 1 | 0 | 0 | 0 | ||||||

1 | 4 | 6 | 4 | 1 | 0 | 0 | ||||||

1 | 5 | 10 | 10 | 5 | 1 | 0 | ||||||

1 | 6 | 15 | 20 | 15 | 6 | 1 | ||||||

1 | 7 | 21 | 35 | 35 | 21 | 7 |

**Figure 11.** Bernoulli’s “Table of Combinations” (Bernoulli, p. 87, or Smith, pp. 86-87)

We would write the expression on the left side of this equation as $${{(n - 1)(n - 2)(n - 3)} \over {1 \cdot 2 \cdot 3}}\quad {\rm or} \quad {n-1\choose 3}.$$

Bernoulli then noted that the sum of all terms in the fourth column up to this term (equivalently, the next term in the fifth column) was $${{n.\,n - 1.\,n - 2.\,n - 3} \over {1.2.3.4}} = {{n^4 - 6n^3 + 11nn - 6n} \over {24}}.$$

Using his summation (integral) notation, he then had

$$\int {n^3 - 6nn + 11n - 6 \over 6} = {n^4 - 6n^3 + 11nn - 6n \over 24}$$

or $$\int {1 \over 6}\,n^3 = {n^4 - 6n^3 + 11nn - 6n \over 24} + \int nn - \int {11 \over 6}\,n + \int 1.$$

Noting that \(\int 1 = n\) and using the formulas for the sums of the integers and their squares given above and then multiplying by 6, Bernoulli obtained the polynomial formula for the sum of the cubes given above. He then noted, “Thus we can step by step reach higher and higher powers . . ..” (Smith, p. 89), and provided a table of formulas for sums of powers up to

$$\int n^{10} = {1 \over 11}\,n^{11} + {1 \over 2}\,n^{10} + {5 \over 6}\,n^9 - 1\,n^7 + 1\,n^5 - {1 \over 2}\,n^3 + {5 \over 66}\,n.$$

From the formulas given above for the sums of the integers, their squares, cubes, and tenth powers, we might conjecture that a formula for the sum of the *c*th powers, where *c* is any positive integer, would have first and second terms $${1 \over {c + 1}}n^{c + 1} + {1 \over 2}n^c.$$

It would be more difficult to see that the remaining powers of *n* are $$n^{c - 1} ,\;n^{c - 3} ,\;n^{c - 5} ,\; \ldots ,\;n$$

for *c* even and $$n^{c - 1} ,\;n^{c - 3} ,\;n^{c - 5} ,\; \ldots ,\;nn$$

for *c* odd, and it would be nearly impossible to see what Bernoulli saw; namely, that the coefficients of these terms involve what soon would become known as Bernoulli numbers. Bernoulli gave the formula

\(\displaystyle\int n^c = {1 \over {c + 1}}n^{c + 1} + {1 \over 2}n^c + {c \over 2}An^{c - 1} + {{c.\,c - 1.\,c - 2} \over {2.3.4}}Bn^{c - 3}\, +\) $${{c.\,c - 1.\,c - 2.\,c - 3.\,c - 4} \over {2.3.4.5.6}}Cn^{c - 5} + {{c.\,c - 1.\,c - 2.\,c - 3.\,c - 4.\,c - 5.\,c - 6} \over {2.3.4.5.6.7.8}}Dn^{c - 7} ... $$

“and so on” (Smith, pp. 88, 90). His instructions for computing the Bernoulli numbers, *A, B, C,* …, were as follows.

The capital letters

A, B, C, Ddenote in order the coefficients of the last terms in the expressions for $$\int nn\ ,\ \int n^4\ ,\ \int n^6\ ,\ \int n^8\ ,$$etc., namely,

Ais equal to 1/6,Bis equal to –1/30,Cis equal to 1/42,Dis equal to –1/30, ...These coefficients are such that each one completes the others in the same expression to unity. Thus

Dmust have the value –1/30 because$${1 \over 9} + {1 \over 2} + {2 \over 3} - {7 \over 15} + {2 \over 9} + ( { + D}) - {1 \over 30} = 1.$$

(Smith, p. 90)

Here, Bernoulli used $$A = {1 \over 6}, \ B = - {1 \over {30}}, \ C = {1 \over {42}},$$

and *c* = 8 in his formula for the sum of the *c*th powers. We would write the sum of the coefficients as $${1 \over 9} + {1 \over 2} + {2 \over 3} - {7 \over {15}} + {2 \over 9} + D = 1$$

and solve for *D,* obtaining *D* = – 1/30. We can see from the formula above for the sum of the 10th powers or by computing directly that the next Bernoulli number, *E*, has the value *E =* 5/66.

With values for the Bernoulli numbers *A, B, C, D,* and *E* and Bernoulli’s formula for the sum of the *c*th powers in hand, we can write formulas for the sums of the 10th and 11th powers and we can compute the next Bernoulli number *F.* This, in turn, would allow us to write formulas for the sums of the 12th and 13th powers and to compute the next Bernoulli number *G,* and so on. Note that if we wanted a formula for, say, the sum of the 43rd powers, we would not have to compute formulas for all the sums of lower degree powers. However, we would have to compute the first 21 Bernoulli numbers and computing the first 21 Bernoulli numbers involves computing the coefficients in the formulas for the 21 sums of the *c*th powers with *c* even and between 2 and 42.

We encourage you to read the passage in which Bernoulli derived formulas for sums of positive integer powers (Smith, pp. 85-90, or Struik, pp. 316-320). The translation from Latin to English in both sources is by Jekuthiel Ginsburg of Yeshiva College (now part of Yeshiva University).

**Exercise 21:** Use Bernoulli’s formula for

$$\int {n^c } \quad ({\rm or}\quad \sum_{k = 1}^n {k^c })$$

to write formulas for $$\int {n^4 } = \sum_{k = 1}^n k^4 , \int {n^5 }=\sum_{k = 1}^n {k^5 }, \quad {\rm and} \quad \int n^{11} = \sum_{k = 1}^n k^{11}.$$

**Exercise 22:** Use Bernoulli’s formula for $$\int {n^c }\quad ({\rm or} \quad \sum_{k = 1}^n {k^c })$$

to write a formula for $$\int n^{12} =\sum_{k = 1}^n k^{12}.$$

Note that you will need to compute the Bernoulli number *F* in order to write this formula.

Solutions to these exercises may be found by clicking here.