已知数列｛an｝,Sn是数列{an}的前n项和,并且满足a1=1,对任意n属于N*,S(n+1)=4an+2
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Sn+1=4an+2
Sn=4a(n-1)+2
a(n+1)=Sn+1-Sn=4an-4a(n-1)
a(n+1)-2an=2an-4a(n-1)
[a(n+1)-2an]/[an-2a(n-1)]=2,为定值.
S2=a2+a1=a2+1=4a1+2=4+2=6
a2=6-1=5
a2-2a1=5-2=3
数列{a(n+1)-2an}是以3为首项,2为公比的等比数列.
a(n+1)-2an=3×2^(n-1)
a(n+1)/2^n-2an/2^n=3×2^(n-1)/2^n
a(n+1)/2^n-an/2^(n-1)=3/2,为定值.
a1/2^(1-1)=1/1=1
数列{an/2^(n-1)}是以1为首项,3/2为公差的等差数列.
an/2^(n-1)=1+(n-1)(3/2)=(3n-1)/2
cn=an/2^n=(1/2)[an/2^(n-1)]=(3n-1)/4
c1=(3-1)/4=1/2
cn-c(n-1)=(3n-1)/4-[3(n-1)-1]/4=1/4,为定值.
数列{cn}是以1/2为首项,1/4为公差的等差数列.
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