(1)1,2,6,24,
120
120

(2)2
1
4
,3
1
9
,4
1
16
,5
1
25
,
6
1
36
6
1
36

(3)觀察下列等式:32-12=8,42-22=12,52-32=16,62-42=20
請(qǐng)猜想:20082-20062=
8028
8028
分析:(1)從第二個(gè)數(shù)起,每一個(gè)數(shù)是前面的數(shù)乘2、3、4、5…所得;
(2)整數(shù)部分是連續(xù)的自然數(shù),分?jǐn)?shù)部分的分母是整數(shù)部分的平方,分子是1;
(3)因?yàn)?2-12=8=2×4,,42-22=12=2×6,52-32=16=2×8,62-42=20=2×10,所以a2-b2(a=b+2)=2×(a+b).
解答:解:(1)24×5=120,

(2)5+1=6,
62=36,
所以應(yīng)該填:6
1
36
,

(3)20082-20062=2×(2008+2006),
=2×4014,
=8028,
故答案為:120;6
1
36
;8028.
點(diǎn)評(píng):根據(jù)給出的數(shù)列,總結(jié)、歸納出變化的規(guī)律,再由規(guī)律解決問題.
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