Question

已知等比數(shù)列{a_n}共有m項(xiàng)（m≥3）,且各項(xiàng)均為正數(shù)，a₁=1,a₁+a₂+a₃=7.

(1)求數(shù)列{a_n}的通項(xiàng)a_n;

(2)若數(shù)列{b_n}是等差數(shù)列，且b₁=a₁,b_m=a_m,判斷數(shù)列{a_n}前m項(xiàng)的和S_m與數(shù)列{b_n-的前m項(xiàng)和T_m的大小并加以證明.

Answer 1

解:(1)設(shè)等比數(shù)列{a_n}的公比為q,則1+?q+q²=7,∴q=2或q=-3,_?

∵{a_n}的各項(xiàng)均為正數(shù)，_?

∴q=2,所以a_n=2^n-1._?

(2)由a_n=2^n-1得S_m=2^m-1.數(shù)列{b_n}是等差數(shù)列，_?

b₁=a₁=1，b_m=a_m=2^m-1,_?

而T_m=(b₁-)+(b₂-)+(b₃-)+…+(b_m-)=(b₁+b₂+b₃+…+b_m)-

=m-m=m·2^m-2,_?

∵T_m-S_m=m·2^m-2-(2^m-1)=(m-4)·2^m-2+1.

∴當(dāng)m=3時(shí)，T₃-S₃=-1,

∴T₃＜S₃，_?

∴當(dāng)m≥4時(shí)，T_m＞S_m.