设等比数列{an}的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数...答:设等比数列{an}的公比为q,前n项和为Sn,且Sn+1,Sn,Sn+2成等差数列,则2Sn=Sn+1+Sn+2,若q=1,则Sn=na1,式显然不成立,若q≠1,则为2 a1(1-qn)1-q = a1(1-qn+1)1-q + a1(1-qn+2)1-q ,故2qn=qn+1+qn+2,即q2+q-2=0,因此q=-2.故答案为-2.
设数列{an}的前n项和为Sn,已知Sn=2an-2n+1,(n为下标,n+1为上标),求通...答:Sn=2an-2n+1,得,a1=2a1-2^2,得a1=4 Sn=2an-2^(n+1),得Sn+1=2an+1-2^(n+2)两式相减,得 an+1=2an+1-2an-2^(n+1)an+1=2an+2^(n+1)两边队以2^(n+1),得 an+1/2^(n+1)=an/2^n+1 an/2^n=a1/2+(n-1)=n+1 所以,an=(n+1)2^n ...