[PDF] X X 2! + t4. 4! 3 t6.





Previous PDF Next PDF



1. 1 + 2 + 3 + ··· + n = n(n + 1)(2n + 1) 6 Proof: For n = 1 the

In Exercises 1-15 use mathematical induction to establish the formula for n ? 1. 1. 1. 2. + 2. 2. + 3. 2. + ··· + n. 2. = n(n + 1)(2n + 1).



MATHEMATICAL INDUCTION SEQUENCES and SERIES

?n is always divisible by 30. Use the fact that n. 5. ?n = (n?1)n(n+1)(n. 2. +1) to prove that it is divisible by 2 and 3 as well as 5.



Problem: For each positive integer n the formula 1 · 3+2 · 4+3 · 5 +

1 · 3+2 · 4+3 · 5 + ··· + n(n + 2) = n(n + 1)(2n + 7). 6 is valid. Proof: (formal style; it is good to do a few proofs this way) We will use the Principle 



Proof by Induction - University of Plymouth

12-Feb-2006 Example 3: for n a natural number prove that: 1) if n ? 2 then n3 ? n is always divisible by 3



X X

2! + t4. 4! 3 t6. 6! + 17 t8. 8!



Problem Solving for Math Competitions Harm Derksen

* Prove that. 12 + 22 + ··· + n2 = n(n + 1)(2n + 1). 6 for all positive integers n. Exercise 1.2. * Show that. 1 ?. 1. 2. +. 1. 3. ? 



BINOMIAL THEOREM

8.1.3 Some important observations. 1. The total number of terms in the binomial expansion of (a + b)n is n + 1 i.e. one more than the exponent n. 2.



CS240 Solutions to Induction Problems Fall 2009 1. Let P(n) be the

If we had shown P(3) as our basis step then the inequality would only be proven for n ? 3. 2. For any positive integer n n. ? i=1. 1 i(i + 1). = 1. 1 · 2.



IMO 2008 Shortlisted Problems

(a ? 1)(b ? 1)(c ? 1) = abc. ?=(a ? 1)2 ? 4a(a ? 1) = (1 ? a)(1 + 3a). ... (i) f(2n ? t(m)) ? 3(n?1)/2 if 2n + m is divisible by 3;.



Untitled

principle of induction P(n) is true for all positive integers n > 1. herefore each term in the product (1+2+22)(1 +3 +3² +3³)(1 +.



Polynomials II - Divisibility and Irreducibles

So now for instance x2 ?2 is not divisible by x? ? 2 over Q since x? ? 2 doesn’t exist over Q Problem 9 Show that x2?2 is irreducible over Q (Hint: Suppose it were reducible What would the degrees of the factors have to be and what does that mean?) Problem 10 Show that x3 ?2 is irreducible over Q (Hint: This is similar



Math 115 Exam  Solutions - Colorado State University

1+n2 = 0 and (ii) the sequence of terms 1+n2 are decreasing To see (i) notice that we can divide numerator and denominator by n2 to get lim n?? 1 n2 ·n 1 n2 (1+n 2) = lim n?? 1 n 1 n2 +1 = 0 To see (ii) let f(x) = x 1+x2 Then f0(x) = (1+x2)·1?x·2x (1+x2)2 = 1?x2 (1+x2)2 ? 0 for x ? 1 Therefore f is a decreasing



3 Mathematical Induction 31 First Principle of

3 MATHEMATICAL INDUCTION 89 Which shows 5(n+ 1) + 5 (n+ 1)2 By the principle of mathematical induction it follows that 5n+ 5 n2 for all integers n 6 Discussion In Example 3 4 1 the predicate P(n) is 5n+5 n2 and the universe of discourse

How to prove n is true for all integers n 1?

Mathematical induction can be used to prove that a statement about n is true for all integers n ? 1. We have to complete three steps. In the basis step, verify the statement for n = 1. In the inductive hypothesis, assume that the statement holds when n = k for some integer k ? 1.

Does f(n) = n 1+n2 converge?

Therefore, f is a decreasing function in the relevant range, so the terms f(n) =n 1+n2are decreasing. 1 We know that the series converges, but we need to determine whether it converges absolutely or not. In other words, we must determine if X? n=1 (?1) nn 1+n2 = X? n=1 n 1+n2 . converges or not.

Is p divisible by Q over C?

Definition 1Let p and q be polynomials in the complex numbers C. We say that p is divisble by q over C if there exists a polynomial r such that p(z) = q(z)r(z). Problem 1 By long division (which we learned how to do last week), answer the following: • Is x3?2x2+ x?2 divisible by x?2 over C?

Ann. Sci. Math. QuÂebec23(1999), no. 1, 63±72.

ON Aq-SEQUENCE THAT GENERALIZES

THE MEDIAN GENOCCHI NUMBERS

G

UO-NIUHAN ANDJIANGZENG

RÂESUMÂE. Dans un article prÂecÂedent [7], nous avons dÂe®ni unq-analogue des nombres de Genocchi mÂedians H 2n+1 . Dans cet article nous dÂemontrons unq- analogue d'unrÂesultat de Barsky [1] sur l'Âetude2-adiquedes nombres de Genocchi mÂedians. A BSTRACT. In a previous paper [7], we de®ned a sequence ofq-median Genocchi numbers H 2n+1 . In the present paper we shall prove aq-analogue of Barsky's theorem about the 2-adic properties of the median Genocchi numbers.

1. Introduction.TheGenocchi numbersG

2n ,n?1, (see [2, 10]), are usually de®ned by their exponential generating function 2 t e t+1 ?t+ Xn? 1 ??1? nG 2n t 2n ?2n?! ?t? t 2 2!+ t 4 4! ?3 t 6 6!+17 t 8 8!

Themedian Genocchi numbersH

2n+1 ,n?0, (see [1, 11]), can be de®ned by H 2n+1 Xi? 0 ??1? iG 2n?2i n 2i+1 ??n? 0?

For exampleH

7 ?3G 6 ?G 4 ?9?1?8. A less classical de®nition of the Genocchi numbers and median Genocchi numbers is the so-called Gandhi generation [3]: G 2n+2 ?Bn ?1??H 2n+1 ?Cn ?1??n?1? whereBn ?x?andCn ?x?,n?1, are polynomialsde®ned by: ?B 1 ?x??x 2 ?B n ?x??x 2 fBn? 1 ?x+1??Bn? 1 ?x?g?n?2? (1) and ?C 1 ?x??1?C n ?x?? ?x +1?f?x+1?Cn? 1 ?x+1??xCn? 1 ?x?g?n?2? (2) RecËu le 29 juillet1997 et, sous forme dÂe®nitive, le 27 fÂevrier 1998. c ?Association mathÂematique du QuÂebec

64On aq-sequence that generalizes the median Genocchi numbers

The reader is referred to [1, 3, 4, 7] for further information about these numbers.

In [7]

1 , we introduced aq-analogue of the above polynomialsby letting ?C 1 ?x? q??1?C n ?x? q?? ?

1+qx??q

?xCn? 1 ?x? q???n?2(3) where ?q f?x?? ?f?1+qx??f?x????1+qx?x?for any polynomialf?x?.As Cn ?x?1??Cn ?x?forn?1, the polynomialsCn ?1?q?,n?1, de®ne aq-analogue of the median Genocchi numbers H 2n+1 In [7], we proved somecombinatorialinterpretationsand analyticalproperties of the polynomial Cn ?1?q?. In this paper, we investigate divisibility properties ofCn ?1?q?.

Our resultsmay be seen as

q-analogues ofthe2-adicpropertiesof themedianGenocchi numbers derived by Barsky [1]. We ®rst recall his result. It will be reproved in section

2 using a different technique.

Theorem 1.(Barsky)For

n?2, the n-th median Genocchi numberH 2n+1 is divisible by2 n?1 and H 2n+1 2 n?1

2?mod 4??ifnis odd?

3?mod 4??ifnis even?

Ourq-analogue of this result is the following theorem.

Theorem 2.The

q-polynomialsC 2n+1 ?1?q?,n?1, andC 2n+2 ?1?q?,n?0,are divisible by ?1+q? 2n+1 . Moreover, forn?1,let c 2n+1 ?q?? C 2n+1 ?1?q? ?1+q? 2n+1 ?c 2n ?q?? C 2n 1?q? ?1+q? 2n?1 ?(4) Then c 2n

1????1?

n +1 G 2n ?c2n+1 ??1?? ??1? n +1 G 2n+2 ?n?1?(5)

The ®rst values of the sequence

fcn ?q?gare the following: c 2 ?q??c 3 ?q??1?c 4 ?q??1+3q+2q 2 +q 3 ?c 5 ?q??1+5q+5q 2 +5q 3 +2q 4 +q 5 We also calculate the ®rst values of the sequencesfH 2n+1 g,fcn ?1?gandfcn ??1?gin the following table: n23 4 5 6 7 8 H 2n+1

2 8 56 608 9440 198272 5410688

cn ?1?1 1 7 19 295 1549 42271 cn ??1?1 1 -1 -3 3 17 -17 1 Preprint available athttp://cartan.u-strasbg.fr/?guoniu/papers

G. Han and J. Zeng65

Note thatH

2n+1 ?2 nc n ?1?or 2 n?1 cn ?1?according to the parity ofn.

Throughoutthispaper,the

q-analogueofnisdenotedby?n?q ?1+q+q 2 +???+q n? 1

Notice that

?n?1 ?n,?

2n+1??

1 ?1and?2n?q

1+q???n?q

2. Flajolet [6] has shown how to derive certain arithmetic properties of a sequence from the continued fraction expansion of its ordinary generating function. Following Flajolet's idea we derive Theorem 1 from the known continued fraction expansion of the median Genocchi numbers (see section 2). Theorem 2 will be proved by a different method (see Section 3). In Section 4, we shall derive some divisibility properties of cn ?q?by continued fraction manipulations that will bring more light in Theorem 2. Finally we make some comments on a possible combinatorial approach to Theorem 2 following a recent work by Kreweras [8] (see Section 5).

2. A new proof of Theorem 1.In thispaper, the continued fractions are considered as

formal power series. Clearly, Cn ?1?q??Cn +1?

0?q?forn?1, so that

1+ Xn? 1 Cn ?1?q?t n? Xn? 1Cn ?0?q?t n? 1 ?(6) In our previous paper [7, Corollary 4], we proved the following identity : Xn?1 Cn ?x? q?t n? t 1?

1+qx??t

1? q?

1+qx??t

1? 2 ?q ???2?q+q 2 x??t 1 q?2?q ???2?q+q 2 x??t ?(7)

It follows from (6) and (7) that

1+

Xn?1Cn

?1?q?t n?1+ t 1? 1?q ??2?q ?t 1? q? 1?q ??2?q ?t 1? 2?q ??3?q ?t 1? q? 2?q ??3?q ?t 1? 3?q ??4?q ?t 1? q? 3?q ??4?q ?t ...(8) 1 100
1? 2 q ?t 1? q? 1? 2 q ?t 1? 2? 2 q ?t 1? q? 2? 2 q ?t 100
3? 2 q ?t 100
q? 3?quotesdbs_dbs35.pdfusesText_40
[PDF] la gestion d'entreprise pdf

[PDF] montrer que n(n+1)(2n+1) est divisible par 3

[PDF] n^3-n est divisible par 6

[PDF] montrer que n(n+1)(n+2) est multiple de 3

[PDF] montrer que n^3-n est divisible par 3

[PDF] montrer que n(n 1)(n 2)(n 3) est divisible par 24

[PDF] الموقع الرسمي للتكوين المهن

[PDF] التكوين المهني بالمغرب

[PDF] ofppt sidi maarouf

[PDF] التسجيل في التكوين المهني

[PDF] ista meknes

[PDF] takwine

[PDF] rapport pisa 2016 france

[PDF] pisa classement

[PDF] pisa 2015 france