Question

已知{an}是各项为正数的等比数列，且a1+a2=(1/a1+1/a2)，a3+a4+a5=64(1/a3+1/a4+1/a5)

已知{an}是各项为正数的等比数列，且a1+a2=(1/a1+1/a2)，a3+a4+a5=64(1/a3+1/a4+1/a5)<1>求{an}的通项公式<2>设bn=(an+1/an)2，求数列{bn}的前n项和Tn

Answer 1

1) 设 a1 = x, 比值为 q x + xq = 2(1/x + 1/(xq)) xq^2 + xq^3 + xq^4 = 64(1/(xq^2) + 1/(xq^3) + 1/(xq^4)) q=2 x=1 an = 2^(n -1) 2) bn = (2^(n-1) + 2^(1-n)) ^2 = 2^(2n-2) + 2 + 2^(2-2n) Tn = (1-2^(2n))/-3 + 2 + (1-2^(-2n))/(3/4) = (2^(2n) - 2^(-2n+2) ) /3 +3