»º º
December 28, 2004
1
1
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
üm
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
.
1.1 (
ª)
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
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