Some new identities in combinatoric

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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.

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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 =− ( x − α1 ) ...( x − α n ) we deduce 1+ x 2 + x n+x ( x + 1) ...( x + n ) x1 ( x + 2 )( x + 3) ...( x + n ) + x2 ( x + 1)( x + 3) ...( x + n ) + x3 ( x + 1)( x + 2 )( x + 4 ) ...( x + n ) + x4 ( x + 1) ...( x + n ) +... + xn ( x + 1)( x + 2 ) ...( x + n − 1) − ( x + 1)( x + 2 ) ...( x + n ) =− ( x − α1 )( x − α 2 ) ...( x − α n ) From straightforward computation, we obtain the solution of the above system equations  n ( −1) ∏ (1 + α ) n −1  i  x1 = i =1 with x = −1  ( n − 1)!  n  ( −1) ∏ ( 2 + α i ) n−2   x2 = with x = −2 i =1  1!( n − 2 )!  n ( −1) ∏ ( 3 + α i ) n −3    x3 = i =1 with x = −3  2!( n − 3) !   ...   n ( −1) ∏ ( n + α i ) n−n   xn = i =1 with x = −n.  ( n − 1)! n ( x) Proposition 2.2. Set 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 + i  n ∑ ( −1)   ( −1) . n (iii) = i =0  i  i  ISSN: 1844 – 9581 Mathematics Section
220 Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh ( −1) ϕ ( i ) n −i n Proof: (i) Similar to Example 2.1, we deduce that: ∑ i =1 ( x + i )( i − 1) !( n − i ) ! −1 ( x − α1 )( x − α 2 ) ... ( x − α n ) n ( −1) ϕ ( i ) − 1 =− ( −1) ϕ ( 0 ) n −i n +1 = − ( x + 1)( x + 2 ) ... ( x + n ) . Substitute x = 0 , we have ∑ i =1 i !( n − i )! n! n n ∑ ( −1)  i ϕ ( i ) = ( −1) n! i n or i =0   n n (ii) Substitute α= ...= α= ∑ ( −1)  i ϕ ( i ) = ( −1) n! we get i n 1 n 0 in i =0   n   n ∑ ( −1)  i  i = ( −1) n!. i n n i =0   n  n  n + i  α i i,= ∑ ( −1)  i  ( −1) . □ i n (iii) Substitute = i 0, 1, . . ., n we obtain = i =0   i  Example 2.3. Solve the following system of linear equations:  n sxk + yk 1 ∑ 2 =  k =1 s + k 2 s s =  ±1, . . ., ± n; xk , yk ∈  ∏(s ) n n 2 + k2 Then, evaluate the sum S = ∑ s =1 . k =1 ∏(k s≠k 2 − s2 1 + k 2)( ) 1 n x x + yk p ( x) Proof: Consider f ( x) =− + ∑ k2 = n , where p ( x) is a x k =1 x + k ( ) 2 x∏ x + k 2 2 k =1 polynomial of the degree ≤ 2n . Since f ( s ) = 0 , we have p ( s ) = 0 where s = ±1, ... , ± n . We obtain f ( x ) = ( ) ( a x 2 − 12 ). )( x 2 − 22 ... x 2 − n 2 x∏ ( x + k ) n 2 2 k =1 1 x x+ y a ( x − 1 )( x − 2 ) ...( x n 2 2 2 2 2 − n2 ) we deduce Since − + ∑ = k k x +k x∏ ( x + k ) 2 2 n x k =1 2 2 k =1 ( ) ( ) ( )( x + 3 ) ... ( x + n ) − x 2 + 12 ... x 2 + n 2 + x ( x1 x + y1 ) x 2 + 22 2 2 2 2 + x ( x x + y ) ( x + 1 )( x + 3 ) ... ( x + n ) + ... + x ( x x + y ) ( x + 1 )( x 2 2 2 2 2 2 2 2 n n 2 2 2 ) ( + 22 ... x 2 + ( n − 1) 2 ) =a ( x − 1 )( x − 2 ) ... ( x − n ) 2 2 2 2 2 2 www.josa.ro Mathematics Section
Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh 221 a =− ( −1) n with x =0   ( ) n a ( −1) ∏ 12 + k 2 n  = − x1 + iy1 = k =1 with x i ( ) n   ∏ k =2 k −1 2  ( ) n  a ( −1) ∏ 22 + k 2 n and obtain the solution  = − x2 + iy2 = k =1 with x 2i ( ) n   ∏ k =2 k −2 2 2  ...  ( ) n ( ) ∏ n2 + k 2 − n  a 1 = − x + iyn = k =1 with x ni  n ( ) n  ∏ k −n 2 2  k =2 ∏(r ) n 2 + k2 Hence y= 1 ...= y= n 0 and xr = k =1 with r = 1, ... , n . ∏ (k ) n 2 −r 2 = k 1, k ≠ r 1 n xk x +∑ 2 ( a x 2 − 12 x 2 − 22 ... x 2 − n 2 )( ) ( ) n xk Since − x k =1 x + k 2 = , we have ∑1+ k = 1 or ( ) n 2 x∏ x + k 2 2 k =1 k =1 ∏(s ) n n 2 + k2 ∑ s =1 = 1 by replacing x = 1 . □ ∏(k )(1 + k ) n k =1 2 −s 2 2 s≠k Proposition 2.4. 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 ∏(s ) n 2 + k2 n ∏ (k ) n ∏ (k − s) n Proof: Since ∑ s =1 = 1 by Example 2.3. and 2 − s2 = ∏ (k )( ) n s ≠ nk s ≠ nk k =1 2 − s2 1 + k 2 s ≠ nk ) ( n 2k 2 ∏ s 2 + k 2 n ( n − k ) !( n + k ) ! , we have n ∏ ( k + s ) = ( −1) ∑ −1 n − k ! n + k ! 1 + k = 1 . n−k s =1 ( ) ( )( )( ) n−k s ≠ nk 2k 2 k =1 2 ISSN: 1844 – 9581 Mathematics Section
222 Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh  2n  ( ) n ( −1) k 2 ∏ s2 + k 2  n−k   n − k  = 2n ! . n By multiplying ( 2n )! we obtain 2∑ s =1 ( ) □ k =1 1+ k2 Example 2.5. Solve the following system of linear equations:  n xk n yk 1 ∑ + ∑ = = s−k k 1 s+k s  k 1= s =  ±i, . . ., ± ni; xk , yk ∈  ∏(s + k ) n 2 2 n Then, evaluate the sum S = ∑ . s =1 k (1 + k ) ∏ ( k − s ) k =1 2 2 2 2 s≠k 1 n x n y p ( x) Proof: Consider f ( x ) =− +∑ k +∑ k = n , where p ( x ) is a x−k k 1 x+k = x k 1= x∏ x − k 2 2 ( ) k =1 polynomial of the degree ≤ 2n . Since f ( s ) = 0 , we have p ( s ) = 0 where s = ±i, ... , ± ni . It is easy to show that f ( x ) = ) (( a x 2 + 12). )( x 2 + 22 ... x 2 + n 2 x∏ ( x − k ) n 2 2 k =1 1 x n y a ( x + 1 )( x + 2 ) ...( x n 2 2 2 2 2 + n2 ) there is: Since − + ∑ +∑ = k k x−k x+k x∏ ( x − k ) n x =k 1=k 1 2 2 k =1 −( x 2 − 12 )...( x 2 − n 2 ) + x1 x( x + 1)( x 2 − 22 )( x 2 − 32 )...( x 2 − n 2 ) + x2 x( x + 2)( x 2 − 12 )( x 2 − 32 )...( x 2 − n 2 ) +... + xn x( x + n)( x 2 − 12 )( x 2 − 22 )...( x 2 − (n − 1) 2 ) + y1 x( x − 1)( x 2 − 22 )( x 2 − 32 )...( x 2 − n 2 ) + y2 x( x − 2)( x 2 − 12 )( x 2 − 32 )...( x 2 − n 2 ) + ... + yn x( x − n)( x 2 − 12 )( x 2 − 22 )...( x 2 − (n − 1) 2 ) = a ( x 2 + 12 )( x 2 + 22 )...( x 2 + n 2 ). ∏( ) ∏( x ) n n ϕ ( x) Put = ψ ( x) x 2 + k 2 and = 2 − k 2 . Then, we obtain: k =1 k =1 www.josa.ro Mathematics Section
Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh 223   a =− ( −1) with x =0 n  ( ) n  a∏ 12 + k 2   x1 = k =1 with x 1 ( ) n  2 ⋅ 12 ∏ 1 − k 2  k =2  ( ) n  a∏ 2 2 + k 2   x2 = k =1 with x 2 ( ) n  2⋅2 ∏ 2 − k 2 2 2  = k 1,k ≠ 2 ...   ( ) n  a ∏ n2 + k 2  xn = k =1 with x n  n −1   2 ⋅ n2 ∏ n2 − k 2 ( ) k =1  ( ) n  a∏ 12 + k 2   y1 = with x = −1 k =1 ( ) n  2 ⋅1 ∏ 1 − k 2 2  k =2  ( ) n  a∏ 22 + k 2 y = k =1 with x = −2  2 ( ) n  2⋅2 ∏ 2 − k 2 2 2  =k 1,k ≠ 2 ...   ( ) n  a ∏ n2 + k 2  yn = k =1 n −1 with x = −n.   2⋅n ∏ n − k 2 ( 2 2 ) k =1 1 n xk From − + ∑ +∑ n yk = a x 2 + 12 x 2 + 22 ... x 2 + n 2 ( )( ) ( ) and xk = yk it x−k k 1 x+k ( ) n x∏ x 2 − k 2 = x k 1= k =1 follows −1 + 2 x ∑ 2 xk = a x 2 + 12 2 n ( )( x + 2 )...( x 2 2 2 + n2 ). k =1 x − k ∏( x − k ) 2 n 2 2 k =1 ∏(s + k ) n 2 2 n By replacing x = i we obtain ∑ s =1 = 1. □ k (1 + k ) ∏ ( k − s ) k =1 2 2 2 2 s≠k ISSN: 1844 – 9581 Mathematics Section
224 Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh Proposition 2.6. 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 ∏(s + k ) n 2 2 n Proof: Since ∑ k 1+ k = 1, s =1 by the Example 2.5 and ( )∏(k − s ) k =1 2 2 2 2 s≠k ( n − k )!( n + k )! we have ∏(k ) n n n − s 2 =∏ ( k − s )∏ ( k + s ) =− ( 1) 2 n−k s≠k s≠k s≠k 2k 2 ( ) n n 2∏ s 2 + k 2 ∑ s =1 = 1. ( −1) ( n − k )!( n + k )!(1 + k 2 ) n−k k =1  2n  ( −1) ∏ ( s 2 + k 2 )  n n−k  n n−k   = 2n ! by multiplying with ( 2n )! . We obtain 2∑ s =1 ( ) □ k =1 1+ k2 Example 2.7. Solve the following system of linear equations and evaluate the following sum:  n sxk + yk 1 ∑ s 2 + k 2 = (i)  k =1 2 2 2 ( s s + 1 s + 22 ... s 2 + n 2 )( ) ( )  s = ±1 ... ± n; xk , yk ∈  n xk (ii) T =∑ . k =1 n + k 2 2 n xk x + yk 1 p ( x) Proof: Consider f ( x ) =∑ k =1 x + k − =n with polynomial ( ) ( ) 2 2 n x∏ x 2 + k 2 x∏ x + k 2 2 =k 1=k 1 p ( x ) of degree ≤ 2n . Because f ( s ) = 0 therefore p ( s ) = 0 by replacing s = ±1 ... ± n . Hence f ( x ) = ( a x 2 − 12 )( x ) ( 2 − 22 ... x 2 − n 2 ). x∏ ( x + k ) n 2 2 k =1 Since ∑ k2 x x + yk n − 1 = ( )( a x 2 − 12 x 2 − 22 ... x 2 − n 2 we obtain ) ( ) k =1 x + k ( ) ( ) 2 n n x∏ x 2 + k 2 x∏ x + k 2 2 =k 1=k 1 www.josa.ro Mathematics Section
Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh 225 −1 + x( x1 x + y1 )( x 2 + 22 )( x 2 + 32 )...( x 2 + n 2 ) + x ( x2 x + y2 ) ( x 2 + 12 )( x 2 + 32 )...( x 2 + n 2 ) + ... + ( xn x + yn ) ( x 2 + 12 )( x 2 + 22 )...( x 2 + (n − 1) 2= ) a ( x 2 − 12 )( x 2 − 22 )...( x 2 − n 2 ) and  ( −1) n −1 = a = with x 0 ( n!) 2   ( ) n  a ( −1) ∏ 12 + k 2 n  1 = − x1 + iy1 = k =1 + n with x i ( ) ( ) n  =k 2=k 2 ∏ k −1 2 2 ∏ k −1 2 2  ( ) n  a ( −1) ∏ 22 + k 2 n  1 = − x2 + iy2 = k =1 + n with x 2i ( ) ( ) n   ∏ k=1, k ≠ 2 k −2 2 2 k= ∏1, k ≠ 2 k −2 2 2   ...   ( ) n ( ) ∏ n2 + k 2 n  a − 1 − xn + iyn 1 = =n −1 k =1 + n with x ni   ∏ k −n 2 2 ( ∏ k −n 2 2 ) ( ) =k 2=k 2 ∏(k ) n 2 + s2 1 We = have yk = 0 , xk s =1 − with k = 1, ... , n and (s ) ∏ (s ) n n ( n!) ∏ 2 2 −k 2 2 −k 2 s= 1, s ≠ k s= 1, s ≠ k obtain the identity: n xk x 1 ( x − 1 )...( x − n ) ( −1) n −1 2 2 2 2 ∑ k =1 x + k − = ( ) ( n!) x∏ ( x + k ) 2 2 n n x∏ x 2 + k 2 2 2 2 = k 1= k 1 n x 1 By x = n there is k ∑n +k = . □ ∏(n ) 2 2 n k =1 2 2 2 n +k k =1 Proposition 2.8. There is the following identity:  n  2  2n  ( )  ( −1) 2k  n + k  k −1 n ∏ k 2 + s2  = ( 2 n )! . ∑  s =1 − 1 ( n!) 1+ k ( ) 2 n k =0   2   ∏ k =0 1+ k2 ISSN: 1844 – 9581 Mathematics Section
226 Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh n xk 1 Proof: By Example 2.7 we have ∑1+ k = or the identity ∏ (1 + k ) 2 n k =1 2 k =1   ( ) n n  ∏ k 2 + s2 1  1 1 ∑   s =1 −  1+ k = ( ) ∏ (s ) ( ) n n 2 n  ( n!) ∏ s − k ∏ 2 k =1 2 2 2 − k2  1+ k2  s=1, s ≠ k s= 1, s ≠ k  k= 1 ( −1) k −1 ∏ (s ) n 1 Since 2 −= 2 k ( n − k )!( n + k )! and T = we obtain ∏ (1 + k ) 2 n = s 1, s ≠ k 2k 2 k =1   ( ) n n  ∏ k 2 + s2 1  1 =T ∑  s =1 − n   1+ k2 ( ) ( ) n ( ) ∏ ∏ 2 k =1  n ! s 2 − k 2 s 2 − k 2   s=1, s ≠ k s=1, s ≠ k    ( ) n n  ∏ k 2 + s2 1  ( −1) 2k 2 k −1 = ∑ s =1 −  k =1  ( n !) ( n − k )!( n + k )! 2 ( n − k )!( n + k )! 1 + k 2    n  2n   n ∏ ( k 2 + s2   )  n + k   ( −1) 2k 2 k −1 = ∑  s =1 − k =1  ( n !) ( 2n )! 2 ( 2 n )!  1 + k 2      n  2  2n  ( () ) k −1 n  ∏ k 2 + s 2  −1 2 k   n+k  ( 2 n )! . Hence ∑  s =1 −1  = □ ( n!) 1+ k2 ( ) 2 n k =1     ∏ k =0 1+ k2 Example 2.9. Suppose that all numbers α1 , α 2 , . . ., α n are different and α i + j ≠ 0 for all i, j = 1, 2, . . ., n . Solve the following system of linear equations:  x1 x2 xn 4 1 + α + 2 + α + ... + n + α = 2α1 + 1  1 1 1  x1 x2 xn 4  + + ... + = 1 + α 2 2 + α 2 n + α 2 2α 2 + 1 ...   x1 x2 xn 4 1 + α + 2 + α + ... + n + α = 2α n + 1 .  n n n www.josa.ro Mathematics Section
Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh 227 x1 x x 4 Proof: Consider f ( x= ) + 2 + ... + n − . We can write f ( x ) as 1+ x 2 + x n + x 2x + 1 p ( x) f ( x) = n , where p ( x ) is a polynomial of degre ≤ n . Since f (α i ) = 0 , we ( 2 x + 1) ∏ ( i + x ) i =1 have p (α i ) = 0 , so we obtain p ( x ) = c ( x − α1 )( x − α 2 ) ...( x − α n ) , c ∈  . Hence ( x + 1)( x + 2 ) ...( x + n ) f ( x) = − ( x − α1 )( x − α 2 ) ...( x − α n ) ( 2 x + 1)( x + 1)( x + 2 ) ...( x + n ) By the equation x1 x x 4 c ( x − α1 )( x − α 2 ) ...( x − α n ) + 2 + ... + n − = 1+ x 2 + x n + x 2 x + 1 ( 2 x + 1)( x + 1)( x + 2 ) ...( x + n ) it induces the following identity (1 + 2 x) [ x1 ( x + 2)( x + 3)...( x + n) + x2 ( x + 1)( x + 3)...( x + n) + x3 ( x + 1)( x + 2)( x + 4)...( x + n) + x4 ( x + 1)( x + 2)...( x + n) +... + xn ( x + 1)( x + 2)...( x + n − 1) ] − 4( x + 1)( x + 2)...( x + n) = c( x − α1 )( x − α 2 )...( x − α n ). Therefore, we obtain the solution of system  n ( ) ∏ (1 + α i ) n −1  c − 1  x1 = i =1 with x = −1  1. ( n − 1)!  n  c ( −1) ∏ ( 2 + α i ) n−2   x2 = with x = −2 i =1  3.1! ( n − 2 ) !  n c ( −1) ∏ ( 3 + α i ) n −3  x = i =1 with x = −3  3  5.2! ( n − 3) !  ...  n c ( −1) ∏ ( n + α i ) n−n  x = i =1 with x = −n  n ( 2n − 1) . ( n − 1)!   n 2i − 1  − 4 ∏ 2 −1 = c = i =1 with x .  n 2α i + 1 □ ( −1) ∏ 2 n  i =1 2 ISSN: 1844 – 9581 Mathematics Section
228 Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh n ( x) Proposition 2.10. Denote ϕ= ∏( x + α ) . i =1 i Then, there are some following identities: n ( −1) i  ϕ ( i )  i  = 4n! + −1 n ϕ 0 . n (i) ∑ i =1 2i − 1 c ( ) ( ) ( −1)  n  n 2n ( n!) i −1 2 n (ii) ∑ i =1  i = 2i − 1  i  ( 2 n )! . x1 x x 4 Proof: From the system of equations + 2 + ... + n − = 1+ x 2 + x n + x 2x + 1 c ( x − α1 )( x − α 2 ) ...( x − α n ) we obtain (i) by replacing x1 , ... , xn and x = 0 . The result (ii) ( 2 x + 1)( x + 1)( x + 2 ) ...( x + n ) is deduced from (i). □ Example 2.11. Solve the following system of equations with variables xk , yk :  n xk n yk 1 ∑ s − k ∑ s + k = + = k 1 =k 1 s s 2 − 12 s 2 − 22 ... s 2 − n 2( )( ) ( )  s = ±i, . . ., ± ni; xk , yk ∈   n  n  ∏ k2 + r2 ( ( n!) 2 )  1 and evaluate the sum ∑  r =1 −  . 1+ k ( ) ( ) n n k =1  2 2 k ∏ k −r 2 2 k 2∏ k 2 − r2  r ≠ k r ≠k  n xk n y 1 p ( x) Proof: We write f ( x ) = ∑ +∑ k − = where x−k k 1 x+k ( ) ( ) n n =k 1= x∏ x − k 2 2 x∏ x − k 2 2 =k 1=k 1 the polynomial p ( x ) ∈  [ x ] is of degree ≤ 2n . Since f ( s ) = 0 , we have p ( s ) = 0 where s= ±i, ... , ± ni . Hence f ( x) = ( a x 2 + 12 )( x ) ( 2 + 22 ... x 2 + n 2 ) x∏ ( x − k ) n 2 2 k =1 ( ) n nn a∏ x 2 + k 2 x y 1 Since ∑ k + ∑ k − =k =1 we deduce x−k k 1 x+k ( ) ( ) n n = k 1= x∏ x 2 − k 2 x∏ x − k 2 2 =k 1=k 1 www.josa.ro Mathematics Section
Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh 229 −1 + x1 x( x + 1)( x 2 − 22 )( x 2 − 32 )...( x 2 − n 2 ) + x2 x( x + 2)( x 2 − 12 )( x 2 − 32 )...( x 2 − n 2 ) + ... + xn x( x + n)( x 2 − 12 )( x 2 − 22 )...( x 2 − (n − 1) 2 ) + y1 x( x − 1)( x 2 − 22 )( x 2 − 32 )...( x 2 − n 2 ) + y2 x( x − 2)( x 2 − 12 )( x 2 − 32 )...( x 2 − n 2 ) + ... + yn x( x − n)( x 2 − 12 )( x 2 − 22 )...( x 2 − (n − 1) 2 ) =a ( x 2 + 12 )( x 2 + 22 )...( x 2 + n 2 ). To find a and xk , yk we replace x = 0, 1, . . . , n and obtain    −1 =a = with x 0  ( n !) 2   ( ) n  a ∏ 12 + k 2 1  x1 = k =1 + with x =1 ( ) ( ) n n  2.1 ∏ 1 − k 2 2 2.1 ∏ 1 − k 2 2 =  k 2=  k 2  ( ) n a∏ 2 2 + k 2  1  x2 =n + with x = k =1 2 ( ) ( ) n  2.2 ∏ 2 − k 2 2 2 2.2 ∏ 2 − k 2 2 2  k= 1, k ≠ 2 k= 1, k ≠ 2  ...  ( ) n  a∏ n 2 + k 2  1  xn = + with x = k =1 n −1 n ( ) ( ) n  2.n ∏ n − k 2 2 2 2.n ∏ n − k 2 2 2 = k 1= k 2  ( ) n  a∏ 12 + k 2  1  y1 = + with x = −1 k =1 ( ) ( ) n n  2.1 ∏ 1 − k 2 2 2.1 ∏ 1 − k 2 2 = k 2= k 2  ( ) n  a∏ 2 2 + k 2  1  y2 = + with x =−2 k =1 ( ) ( ) n n  2.2 ∏ 2 − k 2 2 2 2.2 ∏ 2 − k 2 2 2  k= 1, k ≠ 2 k= 1, k ≠ 2 ...   ( ) n  a∏ n 2 + k 2  yn = 1 k =1 n −1 + with x =−n.  ( ) ( ) n  2.n ∏ n − k 2 2 2 2.n ∏ n − k 2 2 2 = k 1= k 2 ISSN: 1844 – 9581 Mathematics Section
230 Some new identities in combinatoric Dam Van Nhi, Tran Trung Tinh ( ) n n n a∏ x 2 + k 2 xk yk 1 ∑ From + ∑ x−k k 1 x+k − =k =1 and xk = yk it follows ( ) ( ) n n =k 1= x∏ x − k 2 2 x∏ x 2 − k 2 = k 1= k 1 ( ) n n 2 2 a∏ x + k x 1 2x ∑ 2 k 2 − 2 =k =1 . By replacing x = i we obtain k =1 x − k ∏(x ) ( ) n n =k 1=k 1 2 − k2 ∏ x −k2 2  n  n xk ( ) = −1 ( − 1 ) n ( n ! ) n  ∏ k2 + r2 n −1 ( 2 n !) 2 ( )  1 2∑ k =1 1 + k = n or n ∑  r =1 −   1+ k . □ ( ) k =1  2 ( ) ( ) ( ) 2 n n 2 ∏k =1 1+ k 2 = ∏k 1 1+ k 2  k ∏ k −r  r ≠k 2 2 k ∏ k2 − r2 2 r ≠k  Proposition 2.12. For all integer n ≥ 1 we have the following identity: k  2n  4  2n  ( −1)   ( n !)    2  n−k  (  n . ) n n 2∑ ( n !) − ∏ k + r  2 2 = 1+ k ( ) 2 n k =0   r =1 ∏ 1+ k 2 k =0 Proof: By Example 2.11 we obtain the identity  n  n  ∏( k2 + r2 )  1 ( ) ( ) ( ) n −1 2 2 − 1 n ! n ! = ∑  k =1  2 r =1 −  2  1+ k (1 + k ) ( ) ( ) n n n 2 = ∏ k 1 2 k ∏  r ≠ k k 2 − r 2 k 2 ∏ r ≠k k 2 − r  By the other way, ( ) n n n 2k 2 ∏ k 2 − r 2 =2k 2 ∏ ( k − r )∏ ( k + r ) =− ( 1) ( n − k ) !( n + k ) ! n−k r ≠k r ≠k r ≠k  2n  4  2n  ( −1)  ( n !)   k    n − k  = ( )  n  . Since ∏ (0 + r ) = n n n we obtain 2∑ ( n !) − ∏ k 2 + r 2  ( n !) 2 2 2 1+ k ( ) 2 n k =1   r =1 ∏ 1+ k2 k =1 r =1  2n  4  2n  ( −1)  ( n !)   k    n − k  = ( )  n. n n there is 2∑ ( n !) − ∏ k 2 + r 2  □ 2 1+ k ( ) 2 n k =0   r =1 ∏ 1+ k2 k =0 REFERENCES [1] Faddev, D., Sominski, I., Recueil D’Exercices D’Algbre Suprieure, Editions Mir- Moscow, 1977. [2] Rivaud, J., Exercices D’Algbre 1, Paris Librairie Vuibert, 1964. www.josa.ro Mathematics Section