Question

在数列{an}中，已知a1=1/2,a(n-1)-an=1/[(2n)2-1],试写出该数列的前四项，并归纳出它的通项公式。越快越好

Answer 1

1/[(2n)2-1] = 1/[(2n-1)(2n+1)] = 1/2*[1/(2n-1)-1/(2n+1)]
an = a(n-1)-1/2*[1/(2n-1)-1/(2n+1)]
a1 = 1/2 = 1/3 + 1/6
a2 = 1/3 + 1/10
a3 = 1/3 + 1/14
a4 = 1/3 + 1/18
an = 1/3 + 1/[2*(2n+1)]

Answer 2

a2=1/2-1/(4^2-1)=1/2-1/15=13/30
a3=13/30-1/(6^2-1)=13/30-1/35=17/42
a4=17/42-1/(8^2-1)=7/18
an=1/3+1/[2(2n+1)]

Answer 3

0 1/3 4/7 9/11 an=2/4-a[n-2] n有范围 》2