Question

数列[an]中,前n项和sn=n²+1 (1)求数列[an]通项公式 (2)设bn=1/anan+？

Answer 1

a(1)=s(1)=2,
a(n+1) = s(n+1)-s(n) = (n+1)^2 - n^2 = 2n+1 = 2(n+1) - 1.
a(1)=2,
n>=2时,a(n) = 2n-1.
b(1) = 1/[a(1)a(2)] = 1/[2*3] = 1/6.
n>=2时,b(n) = 1/[a(n)a(n+1)] = 1/[(2n-1)(2n+1)] = 1/2[1/(2n-1) - 1/(2n+1)],
t(1) = b(1) = 1/6.
n>=2时,t(n) = b(1) + b(2) + b(3) + ...+ b(n-1) + b(n)
= 1/6 + 1/2[1/3-1/5 + 1/5-1/7 + ...+ 1/(2n-3)-1/(2n-1) + 1/(2n-1)-1/(2n+1)]
= 1/6 + 1/2[1/3 - 1/(2n+1)]
= 1/3 - 1/(4n+2),8,数列[an]中,前n项和sn=n²+1 (1)求数列[an]通项公式 (2)设bn=1/anan+
数列[an]中,前n项和sn=n²+1
(1)求数列[an]通项公式
(2)设bn=1/anan+¹(n∈n+)求数列[bn]前n项和tn