Đang chuẩn bị liên kết để tải về tài liệu:
Some new identities in combinatoric

Không đóng trình duyệt đến khi xuất hiện nút TẢI XUỐNG

In this paper "Some new identities in combinatoric", the binomial numbers m n are very important in several applications and satisfy several number of identities. The purpose of this paper is to introduce a new combinatorial integer m,n j and obtain some algebraic identities by means of double combinatorial argument. | Journal of Science and Arts Year 13 No. 3 24 pp. 217-230 2013 ORIGINAL PAPER SOME NEW IDENTITIES IN COMBINATORIC DAM VAN NHI 1 TRAN TRUNG TINH1 _ Manuscript received15.07.2013 Accepted paper 28.07.2013 Published online 15 09.2013. Abstract. In this paper we introduce some new identities in combinatoric. Keywords Equation Identity Combinatoric. 2010 Mathematics Subject Classification 26D05 26D15 51M16. 1. INTRODUCTION n x Proposition 1.1. Denote ϕ x α . Then there are the following identities i 1 i n n 1 i ϕ i 1 n . i n i i 0 i n n 1 i n 1 n . n ii i 0 i i n n 1 n 1 1 . n iii i 0 i i n 1 ϕ i i i 4n 1 n ϕ 0 . n iv i 1 2i 1 c 1 n n 2n n i 1 2 n v i 1 i 2i 1 i 2 n . Proposition 1.2. There is the following identity 2n n 1 k 2 s2 k 2 n k n k 2n . n 2 s 1 k 0 1 k2 1 Ha Noi National University of Education 136 Xuan Thuy Road Cau Giay Hanoi Vietnam. E-mail damvannhi@yahoo.com. ISSN 1844 9581 Mathematics Section 218 Some new identities in combinatoric Dam Van Nhi Tran Trung Tinh Proposition 1.3. For all integer n 1 there is the following identity 2n 1 s 2 k 2 n n k n n k 2n . 2 s 1 k 1 1 k2 Proposition 1.4. For all integer n 1 there is the following identity n 2 2n k 1 n k 2 s 2 1 2 k n k 2 n . s 1 1 n 1 k 2 n k 0 2 k 0 1 k2 Proposition 1.5. For all integer n 1 we have the following identity 2n 4 2n 1 n k n k n . n n 2 n k 2 r 2 2 1 k 2 n k 0 r 1 1 k2 k 0 2. PROVING SOME NEW IDENTITIES IN COMBINATORIC BY USING THE SYSTEMS OF LINEAR EQUATIONS Example 2.1. Assume that α1 α 2 . . . α n and α i j 0 i j 1 2 . . . n . Solve the following system of linear equations x1 x2 xn 1 α 2 α . n α 1 1 1 1 x1 x2 xn . 1 1 α 2 2 α 2 n α2 . x1 x2 xn 1 α 2 α . n α 1 n n n x1 x x p x Proof Consider f x 2 . n 1 where p x is a 1 x 2 x n x n i x i 1 polynomial of degree n. Since f α i 0 therefore p α i 0 i 1 . n . In view of this result we get f x x α1 x α 2 . x α n . x 1 x 2 . x n www.josa.ro Mathematics Section Some new identities in combinatoric Dam Van Nhi Tran Trung Tinh 219 From x1 x x 2 . n 1

Đã phát hiện trình chặn quảng cáo AdBlock
Trang web này phụ thuộc vào doanh thu từ số lần hiển thị quảng cáo để tồn tại. Vui lòng tắt trình chặn quảng cáo của bạn hoặc tạm dừng tính năng chặn quảng cáo cho trang web này.