www.public4you.fora.pl

px941934

Solving Competition Problems undetermined coefficients Examples

A 0,1,[link widoczny dla zalogowanych],2, ..., n). Yang Hui triangle if the first line of three consecutive terms n ratio of 3:4:5, then there is a positive integer blog, making 3cf, Chi! (N - Chi) 14C (Chi-1)! (Chi + 1)! a = dry '4 C (chi +1)! (one a 1)! i one by one - the February - First +1'. . . f3n a 7k one by one 3, which was i4n a 9 to 5 Chi. f Chih-yi 27. I have a solution of 62 equations. Therefore, in the first 62 lines of Yang Hui triangle will appear next to the number three, and their ratio is 3; 4:5. This commentary questions binomial expansion firmly grasp the general formulas, obtained using a binomial coefficient on the ratio of n, is a binary equations,[link widoczny dla zalogowanych], solutions of this equation can be solutions to this problem. * 46 * School of Mathematical Monthly 2OO6 11 series of 6 seek Example 6 The general formula for a set mouth l 1, al a 2a +, then the general formulas of a, one. (High School Mathematics Olympiad Training Problems 60) solution that a,[link widoczny dla zalogowanych], a b, + Sn. + Tn + k (, t, k is the coefficient to be determined), then a +1 a b, fT1 + (+1). + T (+1) + k a 2b, +2 sn +2 tn +2 k +, that is b, + l A 2b, + (+1). + (T a 2s) n + kt-. f +1-0, f eleven 1, so {t2s 10, solve for {t eleven 2, l take the eleven-o. 1 is one by one 3. fb, = 7, then 16 +: 2 A 2 A 1: ... 7 × 2. Therefore, b a 7 × 2 \, analysis, reasoning demanding. usually on the Recurrence transform into a special series (arithmetic or geometric series) to solve, this method reflects the mathematics of the unknown is known the Idea, but using the undetermined coefficient constant transformation Recurrence is an important transformation method. (Continued from page 41) 6ln, a 6, f +2 oral l-d2 (2 tolerance for the {C}), so ( 5) 6, port +16 + ln a d2. (6) (5) A (6), have b + l (n, +2 + a) a (6 +2 + b,) port +1. (7 ) because the {b} is arithmetic progression,[link widoczny dla zalogowanych], so b, parts + b a 26. and b is not identically 0, so the +2 + a a 2a, +1. (8) so the sequence {a,} is arithmetic series. Analysis of the function point of view of the text I-2-1 by (5), the whole process, \by (6), it is not. because ∈ N, the may not have a 1 ∈ N (when n: 1 时 wrong. [1] in dealing with cases of 2 of the whole process is also guilty of this wrong), but also can not help but fall into a series of \(7) introduction (8). counterexample: take S-. +1, b one by one 2, C, one by one 3 to 4 (to meet the meaning of the questions), but a one {a, sequence {an} is not arithmetic series. explore: (1) Let [2],[link widoczny dla zalogowanych], Theorem l Medium Arithmetic {b}, {C} of tolerance, respectively d, d., by the given recurrence formula, we can see when a l,, bl (mouth 1 + a2) a 63 l: =: C1. which may bla2 a b2al = a a1 (b1 + b2 ~ b3) + C1 eleven a1 (61 A d) + f.. To make (6) of the have set up any ∈ N, as long as === 1 (6) holds can be. For this reason, we can add conditions: a1 (b1 a d1) === c1 a d2; (2) in the conditions \-d1) a Cl-d2 \[2] Theorem 1 corrected as follows: Theorem given numbers out {a) of the preceding paragraph and for the-S series {b,}, {C,}, respectively, for the tolerance is equal to d, d.'s arithmetic series, and b, ... (∈ N) is not constant 0,6,, a 6 +2 S = C (∈ N), and 1 (b1 a d1) a C1 a d., then the sequence {a} for the arithmetic sequence.
More articles related to topics：

[link widoczny dla zalogowanych]

mbt schuhe koeln Protect the children of migrant w

[link widoczny dla zalogowanych]