解:(1)a
1=
,
a
2=
=2,
a
3=
=-1,
a
4=
=
;
(2)∵2010=3×670,
∴a
2010=a
3=-1,a
2011=a
1=
,a
2012=a
2=2,
∴a
2010•a
2011•a
2012=-1×
×2=-1;
(3)∵a
1•a
2•a
3=a
4•a
5•a
6=…a
2008•a
2009•a
2010=
×2×(-1)=-1,
∴a
1•a
2•a
3…a
2010•a
2011•a
2012.=(a
1•a
2•a
3)•(a
4•a
5•a
6)…(a
2008•a
2009•a
2010)•a
2011•a
2012=(-1)
670•
•2=-1.
分析:(1)根據(jù)差倒數(shù)的定義可計(jì)算出a
2=
=2,a
3=
=-1,a
4=
=
;
(2)根據(jù)(1)計(jì)算結(jié)果得到從a
1開(kāi)始,每三個(gè)數(shù)一循環(huán),而2010=3×670,則a
2010=a
3=-1,a
2011=a
1=
,a
2012=a
2=2,然后計(jì)算a
2010•a
2011•a
2012的值;
(3)由于a
1•a
2•a
3=a
4•a
5•a
6=…a
2008•a
2009•a
2010=-1,把a(bǔ)
1•a
2•a
3…a
2010•a
2011•a
2012分成(a
1•a
2•a
3)•(a
4•a
5•a
6)…(a
2008•a
2009•a
2010)•a
2011•a
2012,然后代值計(jì)算即可.
點(diǎn)評(píng):本題考查了規(guī)律型:數(shù)字的變化類:通過(guò)從一些特殊的數(shù)字變化中發(fā)現(xiàn)不變的因素或按規(guī)律變化的因素,然后推廣到一般情況.