已知数列{an}满足a1=1,an-a(n+1)=ana(n+1),数列{an}的前n项和为Sn.(1)求证:{1/an}为等差数列;(2)设Tn=S2n-Sn,求证T(n+1)>Tn

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已知数列{an}满足a1=1,an-a(n+1)=ana(n+1),数列{an}的前n项和为Sn.(1)求证:{1/an}为等差数列;(2)设Tn=S2n-Sn,求证T(n+1)>Tn
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已知数列{an}满足a1=1,an-a(n+1)=ana(n+1),数列{an}的前n项和为Sn.(1)求证:{1/an}为等差数列;(2)设Tn=S2n-Sn,求证T(n+1)>Tn
已知数列{an}满足a1=1,an-a(n+1)=ana(n+1),数列{an}的前n项和为Sn.(1)求证:{1/an}为等差数列;
(2)设Tn=S2n-Sn,求证T(n+1)>Tn

已知数列{an}满足a1=1,an-a(n+1)=ana(n+1),数列{an}的前n项和为Sn.(1)求证:{1/an}为等差数列;(2)设Tn=S2n-Sn,求证T(n+1)>Tn
an-a(n+1)=ana(n+1) 【两边同除以ana(n+1)】
得:
1/[a(n+1)]-1/[a(n)]=1
即:
数列{1/(an)}是以1/a1=1为首项、以d=1为公差的等差数列.
则:
1/[a(n)]=n
a(n)=1/n