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December 28, 2004

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(1)

   

 

  »º º

December 28, 2004

    

(2)

1

 

1

(3)

1

 ª 

  

 n j



=

 n n − j



= n!

j!(n − j)!



n



j

 



(y + 1) n



(y + 1) n =  n

j=0

 n j

 y j .

        

        

   

Ë

1.1 (

 

)



n, m

  

n ≥ m ≥ 1



 m j=0

 n j

 m j



=

 n + m m

 .

  

n + m



m

  



j



 n

j

 m

j

 =  n

j

 m

m−j





n



j



m





m − j

 ü 

m

  

 

 

f (y) = (y + 1) m (y −1 + 1) n

  





(y + 1) m

 

y j

 

 m

j





(y −1 + 1) n

 

y −j

 

 n

j





f (y)

  

 m j=0

 n j

 m j

 .

 

f (y)



f (y) = (y + 1) m (y −1 + 1) n = y m (y −1 + 1) n+m .



f (y)

  



n + m m

 .

 

 m j=0

 n j

 m j



=

 n + m m

 .



(4)



1.1 (



)



n, m

  

n ≥ m ≥ 1



 m j=0

 m j

 2 

n + 2m − j 2m



=

 n + m m

 2 .

 

f (x, y) = (x + 1) n+m (x + 1 + y) m (y −1 + 1) m

 

x 2m

 



(x + 1) n+m (x + 1 + y) m

 

x 2m y j

 

 m j

 n + 2m − j 2m

 ,

(y −1 + 1) m

 

y −j

 

 m

j





f (x, y)

 

x 2m





 m j=0

 m j

 n + 2m − j 2m

 m j



=  m

j=0

 m j

 2 

n + 2m − j 2m

 .



(x + 1 + y) m (y −1 + 1) m =

 m j=0

 m j



x j (1 + y) m−j (y −1 + 1) m

=  m

j=0

 m j



x j y m−j (y −1 + 1) 2m−j ,



(x + 1 + y) m (y −1 + 1) m

 

x j

 

 m

j

 2m−j

m−j





(x + 1) n+m





x 2m−j

 

 n+m

2m−j





f (x, y)

 

x 2m

 

 m j=0

 m j

 2m − j m − j

 n + m 2m − j



=

 n + m m

  m

j=0

 n m − j

 m j



=

 n + m m

  m

j=0

 n m − j

 m m − j



=

 n + m m

  m

j=0

 n j

 m j



=

 n + m m

 2

   

.



f (x, y)

 

x 2m

 



       

   



1.1

 

n

 

n

   

(5)



1.2

 

r, n



1 ≤ r ≤ n



 r r

 +

 r + 1 r

 +

 r + 2 r



+ · · · +

 n r



=

 n + 1 r + 1

 .



1.3



n

    

(a, b, c, d)



0 ≤ a ≤ b ≤ c ≤ d ≤ n.



1.4

 

 n

2

 2



= 3

 n + 1 4



, n ≥ 3,

   



 

n

 

n ≥ 2

  







p

 

n

  



1 − 1729 p n

   

(p, n)



參考文獻