440 



The equation 



5(0, n) x n — 5(1, n) x" l + . . . . + (—1)" B(n, n) = 



k = 1, 2, 3 



has 1, 2, 3, n-, for roots. If we set 



S k = V ■ + 2 K + 3* + + n 



and solve Newton's formulae* we obtain 



Si 1 



S 2 & 2 



Oj >J2 Ol O 



B(Jfc,Jb) B(k,n) = 



Oj O3 0-2 Ol 



*Sfc 'Sfjfc — i 'Sfc— 2 *5a— 3 

 This determinant vanishes when k > n. 

 Inversely, 



I 5(l,n) 5(0,n) 



2B(2,n) B(l,n) B{0,n) 



3B(3,n) B(2,n) B{l,n) . 

 S, 



S, 



k,n = 1, 2, 3. 



JfcB(ft,n) B(k—l,n) B(k—2,n) B(l,n) 



k, n = 1, 2, 3 (even if k > n) 



These sums of the powers of the first n natural numbers are connected 

 by the following relations, in which l{k/2) signifies the integral part of k/2: 



Kk 2) 

 V I * I o _ +-/-1 o * 



-" 2i + lJ "2A— l— 2i— T- ^i 

 i = *■ 



v 2 2k + 1 - 2i {k} „ _f 9ri . n o*-ie* 



whence 



k 

 ; = 



— - c,- • Soi—i= where c,= — - — : — when i is even 



. n l ( l ) M ^ - l+i 



1 



— (2n+l) when i is odd 



*See, for example, Cajori's Theory of Equations, pp. 85-S6. 

 tStern, Crelle's Journal, Vol. 84, pp. 216-218. 



