a1=k/3+1/2,a2=k/6+1/2-k/3-1/2=-k/6,a3=k/9+1/2-k/6-1/2=-k/18,a2^2=a1a3,(-k/6)^2=(k/3+1/2)(-k/18),得k=0或-3/5,k=0,则sn=1/2,不和题目,故k=-3/5,sn=1/2-1/5n,an=sn-sn-1=1/5(n-1)n.n>=2,n=1时a1=3/10.
已知等比数列an中,sn为数列an的前n项和,满足sn=k·1/3n+1/2(1)求k的值及an,(2)设bn=2n+1
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