Question

数列1.在数列{an}中,a1=1,且满足an-an-1=n(n>1)

Answer 1

解：
(1)
a2-a1=2,
a2=a1+2=3,
a3-a2=3,
a3=a2+3=6,
an-a(n-1)=n
an=a(n-1)+n
=a(n-2)+n+(n-1)
=...
=a1+n+(n-1)+...+2
=1+2+...+n
=n(n+1)/2
a1=1也满足此式，
因此通项是an=n(n+1)/2
(2)
bn=1/an
=2/[n(n+1)]
=2/n-2/(n+1)
Sn=b1+b2+..+bn
=(2/1-2/2)+(2/2-2/3)+...+(2/n-2/(n+1))
=2/1-(2/2-2/2)-(2/3-2/3)-...-(2/n-2/n)-2/(n+1)
=2-2/(n+1)