设数列{an}的前n项和Sn,若Sn=a1(3^n—1)/2,且a4=54,则a1=
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设数列{an}的前n项和Sn,若Sn=a1(3^n—1)/2,且a4=54,则a1=
设数列{an}的前n项和Sn,若Sn=a1(3^n—1)/2,且a4=54,则a1=
设数列{an}的前n项和Sn,若Sn=a1(3^n—1)/2,且a4=54,则a1=
a4=S4-S3=a1(3^4-1)/2-a1(3^3-1)/2=27a1;
由于a4=54,所以27a1=54,a1=2
n>1时 an=sn-s(n-1)=a1(3^n-3^(n-1))/2
a4=a1(81-27)/2=27a1=54 ==>a1=2
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