数列题2道1.已知{an}的首项a1=2/3,an+1(+1是在底下的)=2an/an+1(+1在底下)n=1.2.3.求数列{n/an}前n项和S
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数列题2道1.已知{an}的首项a1=2/3,an+1(+1是在底下的)=2an/an+1(+1在底下)n=1.2.3.求数列{n/an}前n项和S
数列题2道
1.已知{an}的首项a1=2/3,an+1(+1是在底下的)=2an/an+1(+1在底下)n=1.2.3.求数列{n/an}前n项和S
数列题2道1.已知{an}的首项a1=2/3,an+1(+1是在底下的)=2an/an+1(+1在底下)n=1.2.3.求数列{n/an}前n项和S
由已知,得:a[n+1]²=2a[n],
两边取e的对数,的lna[n+1]²=ln2a[n],即 2lna[n+1]=lna[n]+ln2,
从而有:lna[n+1]-ln2=1/2(lna[n]-ln2),
所以数列lna[n]-ln2是以lna[1]-ln2为首项,1/2为公比的等比数列.
即 lna[n]-ln2=(lna[1]-ln2)*(1/2)ˆn-1 ①
把a[1]=2/3代入①式整理,得:
lna[n]=ln2-(1/2)ˆn-1*ln3
∴a[n]=2/3ˆ(1/2)n-1