Bernoulli Numbers

ABOUT BERNOULLI’S NUMBERS Prof. Mihàly Bencze Department of Mathematics Áprily Lajos College Braşov, Romania Florentin Smarandache, Ph D Full Professor Chair of Department of Math & Sciences University of New Mexico 200 College Road Gallup, NM 87301, USA E-mail: [email protected]

Abstract. Many methods to compute the sum of the first n natural numbers of the same powers (see [4]) are well known. In this article we present a simple proof of the method from [3]. 2000 MSC: 11B68 Introduction. The Bernoulli’s numbers are defined by −1 (1) Bn = Cn0+1 B0 + Cn1+1 B1 + ... + Cnn+−11 Bn −1 ) ( n +1 where B0 = 1 . It is known that Bn +1 = 0 if n ≥ 1 . By calculation we find that: 1 1 1 1 1 5 B1 = − , B2 = , B4 = − , B6 = , B8 = − , B10 = , (2) 2 6 30 42 30 66 691 7 3617 43867 174611 B12 = − , B14 = , B16 = − , B18 = , B20 = − , 2730 6 510 798 330 854513 236364091 , etc. B22 = , B24 = − 138 2730 Let Snk = 1k + 2 k + ... + n k the sum of the first n natural numbers which have the same power. Theorem. 1 ⎛ k +1 1 1 k ⎞ k −1 2 k (3) Snk = ⎜⎝ n + Ck +1n + Ck +1 B2 n + ... + Ck +1 Bn n ⎟⎠ k +1 2 Proof: (1) can be written as:

1

n

∑C

(4)

i n +1

Bi = 0, n ≥ 1 .

i=0

k

If P(x) = ∑ Cki +1 Bi x k +1−i , i=0

then k ⎛ k +1−i ⎞ − n k +1−i = ∑ Cki +1 Bi ⎜ ∑ Ckj+1−i n k +1−i − j ⎟ . i =0 i =0 ⎝ j =1 ⎠ k −1 Let At be the coefficients of n , where t ∈ {0,1,..., k} . k

P (n + 1) − P(n) = ∑ Cki +1 Bi

(( n + 1)

k +1−i

)

t t ⎞ t +1 i +1 ⎛ i At = ∑ Cki +1Cki ++1− B = C i i k +1 ⎜ ∑ Ci +1 Bi ⎟ . ⎝ i=0 ⎠ i=0

If n ≥ 1 , then At = 0 , only A0 = Ck1 +1 . Because of these P(n + 1) − P(n) = Ck1 +1n k . Using this n −1

1 n −1 1 P(n) , (P(i + 1) − P(i)) = ∑ k + 1 i=0 k +1 i=0 1 because P(0) = 0 . Then Snk = P(n) + n k . From here one obtains (3). k +1 Note. From the previous result we can also find the formula 1 Snk = P(n + 1) . k +1 Using this, we find the following equalities: 1 1 1 2 S n0 = n, S n1 = n ( n + 1) , Sn2 = n ( n + 1)( 2n + 1) , Sn3 = n 2 ( n + 1) , 2 6 4 1 1 2 S n4 = n ( n + 1)( 2n + 1) ( 3n 2 + 3n − 1) , S n5 = n 2 ( n + 1) ( 2n 2 + 2n − 1) , 30 12 1 S n6 = n ( n + 1)( 2n + 1) ( 3n 4 + 6n3 − 3n + 1) , 42 1 2 2 S n7 = n ( n + 1) ( 3n 4 + 6n3 − n 2 − 4n + 2 ) , 24 1 S n8 = n ( n + 1)( 2n + 1) ( 5n6 + 15n5 + 5n 4 − 15n3 − n 2 + 9n − 3) , 90 1 S n9 = ( 2n10 + 10n9 + 15n8 − 14n 6 + 10n 4 − 3n 2 ) , 20 1 S n10 = ( 6n11 + 33n10 + 55n9 − 66n 7 + 66n5 − 33n3 + 5n ) , 66 1 S n11 = ( 2n12 + 12n11 + 22n10 − 33n8 + 44n6 − 33n 4 + 10n 2 ) , 24 1 S n12 = 210n13 + 1365n12 + 3630n11 − 4935n9 + 115n8 + ( 2730 +9640n 7 + 1960n 6 − 5899n 5 + 35n 4 + 4550n 3 + 1382n 2 − 691n ) , etc.

∑ik =

2

Problems: 1) Using the mathematical induction on the base of (1), we prove that B2 n +1 = 0, if n ≥ 1 .

2) Prove that Snk is divisible by n(n + 1) . 3) Prove that Sn2 k +1 is divisible by n 2 (n + 1)2 . 4) Determine those natural numbers n, k n(n + 1)(2n + 1) . 5) Detach in parts the sums Sn9 , Sn10 Sn11 , Sn12 . 6) Using (2), (3), compute the sums Sn13 ,..., Sn21 .

for which Sn2 k

is divisible

REFERENCES

[1] [2] [3] [4] [5]

M. Kraitchik – Recherches sur la théorie des nombres – Paris, 1924. Mihály Bencze – Osszegekrol – A Matematika Tanitasa, 1/1983. Z. I. Borevici, I. R. Safarevici – Teoria numerelor – Ed. Ştiinţifică şi Enciclopedică, Bucharest, România, 1985. Sándor József – Veges osszegekrol – Matematikai Lapok, Cluj-Napoca, 9/1987, România. Mihály Bencze – A Bernoulli szamok egyik alkalmazasa – Matematikai Lapok, Kolozsvar 7/1989, pp. 237-238, România.

[Published in “Octogon Mathematical Journal”, Braşov, Vol. 7, No. 1, 151-3, 1999.]

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