Xem mẫu
- 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
nguon tai.lieu . vn
