已知数列{an}中,Sn是它的前n项之和,并且S(n+1)=4an+2(n=1,2,…)a1=1. 设bn=a(n+1)-2an (n=1,2,…),则数列{bn}是等比数列; 设cn= (n=1,2,…),则数列{cn}是等差数列 ,an=2^n-1 +(n-1) 3*2^n/4,Sn=2+3(n-1)2^n
There is a sequence {an}. Sn is the sum of the first n items of {an}, and S(n+1)=4an+2. a1=1. Then an=2^n-1 +(n-1) 3*2^n/4, and Sn=2+3(n-1)2^n. Suppose bn=a(n+1)-an, then {bn} is a geometric sequence.
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