证明：
∣
a
2
(
a
+
1
)
2
(
a
+
2
)
2
(
a
+
3
)
2
b
2
(
b
+
1
)
2
(
b
+
2
)
2
(
b
+
3
)
2
c
2
(
c
+
1
)
2
(
c
+
2
)
2
(
c
+
3
)
2
d
2
(
d
+
1
)
2
(
d
+
2
)
2
(
d
+
3
)
2
∣
=
0
|a2(a+1)2(a+2)2(a+3)2b2(b+1)2(b+2)2(b+3)2c2(c+1)2(c+2)2(c+3)2d2(d+1)2(d+2)2(d+3)2| = 0
a2b2c2d2(a+1)2(b+1)2(c+1)2(d+1)2(a+2)2(b+2)2(c+2)2(d+2)2(a+3)2(b+3)2(c+3)2(d+3)2
=0
∣ a 2 ( a + 1 ) 2 ( a + 2 ) 2 ( a + 3 ) 2 b 2 ( b + 1 ) 2 ( b + 2 ) 2 ( b + 3 ) 2 c 2 ( c + 1 ) 2 ( c + 2 ) 2 ( c + 3 ) 2 d 2 ( d + 1 ) 2 ( d + 2 ) 2 ( d + 3 ) 2 ∣ = c 2 − c 1 c 3 − c 1 c 4 − c 1 ∣ a 2 2 a + 1 4 a + 4 6 a + 9 b 2 2 b + 1 4 b + 4 6 b + 9 c 2 2 c + 1 4 c + 4 6 c + 9 d 2 2 d + 1 4 d + 4 6 d + 9 ∣ = c 3 − 2 c 2 c 4 − 3 c 2 ∣ a 2 2 a + 1 2 6 b 2 2 b + 1 2 6 c 2 2 c + 1 2 6 d 2 2 d + 1 2 6 ∣ = 0 |a2(a+1)2(a+2)2(a+3)2b2(b+1)2(b+2)2(b+3)2c2(c+1)2(c+2)2(c+3)2d2(d+1)2(d+2)2(d+3)2| \xlongequal{c2−c1c3−c1c4−c1} |a22a+14a+46a+9b22b+14b+46b+9c22c+14c+46c+9d22d+14d+46d+9| \xlongequal{c3−2c2c4−3c2} |a22a+126b22b+126c22c+126d22d+126| = 0 a2b2c2d2(a+1)2(b+1)2(c+1)2(d+1)2(a+2)2(b+2)2(c+2)2(d+2)2(a+3)2(b+3)2(c+3)2(d+3)2 c2−c1c3−c1c4−c1 a2b2c2d22a+12b+12c+12d+14a+44b+44c+44d+46a+96b+96c+96d+9 c3−2c2c4−3c2 a2b2c2d22a+12b+12c+12d+122226666 =0
得证。