数列{an}的前n项和为Sn,Sn=(m+1)-man对于任意的n都成立,其中m为常数,且m不等于1 (1)求

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Answer 1

Sn=(m+1)-man对于任意的n都成立,其中m为常数,且m不等于-1令n=1代入,S1=a1=(m+1)-ma1 → a1=1令n=2代入,S2=a1+a2=(m+1)-ma2 → a2=m/(+1m)令n=3代入,S3=a1+a2+a3=(m+1)-ma3 → a3=[m/(m+1)]^2(1)求证：数列{an}是等比数列S(n)-S(n-1)=(m+1)-man-[(m+1)-ma(n-1)]an/a(n-1) = m/(m+1)所以,数列{an}是以公比q=m/(m+1)的等比数列2)记数列{an}的公比为q,设q=f(m),若数列{bn}满足：b1=a1,bn=f(bn-1)(n=2),求证：{1/bn}是等差数列；b1=a1=1bn=f(bn-1)=b(n-1)/[b(n-1) + 1]1/bn =1 /b(n-1) + 11/bn - 1 /b(n-1) = 1所以,：{1/bn}是以公差d=1的等差数列3)在(2)的条件下，设cn=bn*bn+1,数列{cn}的前n项和为Tn,求证：Tn<1.1/bn = n → bn = 1/ncn=bn*b(n+1)= 1/[n(n+1)] = 1/n - 1/(n+1)