设数列{an}的前n项和Sn,且a1=1,Sn=4an+2(n∈N*)1.设bn=an+1-2an,求证:{bn}是等比数列2.设Cn=an/2∧n,求证:{Cn}是等差数列3.求Sn=a1+a2+...+an

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设数列{an}的前n项和Sn,且a1=1,Sn=4an+2(n∈N*)1.设bn=an+1-2an,求证:{bn}是等比数列2.设Cn=an/2∧n,求证:{Cn}是等差数列3.求Sn=a1+a2+...+an
设数列{an}的前n项和Sn,且a1=1,Sn=4an+2(n∈N*)
1.设bn=an+1-2an,求证:{bn}是等比数列
2.设Cn=an/2∧n,求证:{Cn}是等差数列
3.求Sn=a1+a2+...+an

设数列{an}的前n项和Sn,且a1=1,Sn=4an+2(n∈N*)1.设bn=an+1-2an,求证:{bn}是等比数列2.设Cn=an/2∧n,求证:{Cn}是等差数列3.求Sn=a1+a2+...+an
题目中应该是:S(n+1)=4an+2
∵S(n+1)=4an+2
Sn=4a(n-1)+2
∴S(n+1)-Sn=4an-4a(n-1)
a(n+1)=4an-4a(n-1)
a(n+1)-2n=2(an-2a(n-1))
a(n+1)-2n/(an-2a(n-1))=2
∵bn/b(n-1)=a(n+1)-2n/(an-2a(n-1))=2
∴{bn}是等比数列
S2=4a1+2+6 a1=1 a2=5
b1=a2-2a1=3
∴bn=3*2∧(n-1)=a(n+1)-2an
a(n+1)/2∧(n+1)-an/2∧n=3/4
∴cn=1/2+(n-1)3/4=3n/4-1/4
∴{cn}是等差数列
Sn=a1+a2+...+an
an =[2∧(-2n)]*(3n-1)
在用等差*等比的公式计算就可以了

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