a
设数列{an}的前n项和为Sn,Sn= a1(3n−1) 2(对于所有n≥1), 则a4=S4-S3= a1(81−1) 2− a1(27−1) 2=27a1, 且a4=54,则a1=2 故答案为2
设数列{an}的前n项和为Sn,Sn=a
设数列{an}的前n项和为Sn,且Sn=2^n-1.
设数列{an}的前n项和为Sn,已知a1=a,an+1=Sn
设数列{an}的前n项和为Sn,a1=10,a(n+1)=9Sn+10
设数列{an}前n项和为Sn,数列{Sn}的前n项和为Tn,满足Tn=2Sn-n2,n∈N*.
设数列{an}的前n项和为Sn=2an-2n,
设数列{an}的前n项和为Sn,且对任意正整数n,an+Sn=4096
设 数列{an}的前n项和为Sn,已知b*an - 2^n=(b-1)Sn
设数列{an}的前n项和为Sn,且对任意正整数n,an+Sn=4096.
设数列{an}的前N项和为Sn,已知1/Sn+1/S2+1/S3+.+1/Sn=n/(n+1),求Sn
设数列{an}的前n项和为Sn.已知a1=a,an+1=Sn+3n,n∈N*.由
一道数学题:设数列{an}的前n项和为Sn.已知a1=a,a(n+1)=Sn+3^n,n属于N*.
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