已知数列{an}是各项均不为0的等差数列,Sn为其前n项和,且满足an^2=S2n-1.令bn=1/anxan+1数列{bn}的前n项和为Tn...求数列{an}的通项公式及数列{bn}的前n项和Tn

Question

已知数列{a_n}是各项均不为0的等差数列,S_n为其前n项和,且满足a_n^2=S_2n-1.令bn=1/a_n x a_n+1数列{b_n}的前n项和为T_n...求数列{a_n}的通项公式及数列{b_n}的前n项和T_n

Answer 1

a_n=2n-1,T_n=n/(2n+1) ,算a_n,利用等差数列的公式,a_n^2=S_2n-1,a_n=a_1+(n-1)d,S_2n-1=(2n-1)a_1+(2n-1)(2n-2)d/2,将a_n^2和S_2n-1都关于n整理,左右两边相等,b_n=「(1/2n-1)-(1/2n+1)」×1/2,用列项分解法求和