各项均为正的数列{an}的前n项和为Sn,已知Sn=1/8(an+2)^2各项均为正的数列{an}的前n项和为Sn,已知Sn=1/_数学解答

各项均为正的数列{an}的前n项和为Sn,已知Sn=1/8(an+2)^2
各项均为正的数列{an}的前n项和为Sn,已知Sn=1/8*(an+2)^2,(1)求证：数列{an}是等差数列；(2)数列{bn}满足bn=(t-1)^[(an+2)/4](t>1),Tn为数列{bn}的前n项和,求limTn.

人气：281 ℃　时间：2020-09-05 14:21:18

解答

1.a = S - S= (1/8) * (a + 2)^2 - (1/8)*(a +2)^2= (1/8)*a^2 + (1/2)*a - (1/8)*a^2 - (1/2)*a移项0 = (1/8)*a^2 - (1/2)*a - (1/8)*a^2 - (1/2)*a分解因式(1/8)*[a^2 - a^2] - (1/2)*(a + a) = 0(1/4)*(a + a)(a-...