已知数列{an}的前n项和为Sn,且S(n+1)=4an+2,a1=1,设Cn=an/2^n,求证数列{Cn}是等差数列
人气：460 ℃ 时间：2020-06-09 03:14:42
解答
S(n+1)=4an+2Sn=4a(n-1)+2a(n+1)=S(n+1)-Sn=4(an-a(n-1))a(n+1)-2an=2[an-2a(n-1)][a(n+1)-2an]/[an-2a(n-1)]=2{an-2a(n-1)}是公比为2的等差数列a1=1S2=4+2=6,a2=S2-a1=6-1=5an-2a(n-1)=[a2-2a1]*2^(n-2)=3*2^(n-2)a...
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