Question

已知等差数列{a¬n}中, 首项a1=1，公差d为整数，且满足a1+3＜a3，a2+5＞a4.数列{bn}满足bn=1/

已知等差数列{a­n}中, 首项a1=1，公差d为整数，且满足a1+3＜a3，a2+5＞a4.数列{bn}满足bn=1/ a­n* a­n+1,其前n项和为Sn
(1)求数列{a­n}的通项公式a­n
(2) 求数列{bn}前n项和为Sn

Answer 1

a(n) = 1 + (n-1)d,
a(1)+3 = 4 < a(3) = 1 + 2d, d > 3/2.
a(2)+5=1+d+5>a(4)=1+3d, d < 5/2.
3/2 < d < 5/2. d = 2.

a(n) = 1 + 2(n-1) = 2n-1.

b(n) = 1/[a(n)a(n+1)] = 1/[(2n-1)(2n+1)] = (1/2)[1/(2n-1) - 1/(2n+1)],
s(n) = b(1)+b(2)+...+b(n-1)+b(n)
= (1/2)[1/1 - 1/3 + 1/3-1/5 + ... + 1/(2n-3) - 1/(2n-1) + 1/(2n-1) - 1/(2n+1)]
= (1/2)[ 1 - 1/(2n+1)]
= n/(2n+1)