Question

数列{an}满足关系式a(n+1)=an+n,设bn=1/[a(n+1)-a1],数列{bn}的前n项和为Sn,则S10

Answer 1

a(n+1)=an+n

a(n+1)-an=n

an-a(n-1)=n-1

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

…………

a2-a1=1

累加

a(n+1)-a1=1+2+...+n=n(n+1)/2

bn=1/[a(n+1)-a1]=2/[n(n+1)]=2[1/n -1/(n+1)]

Sn=b1+b2+...+bn=2[1/1-1/2+1/2-1/3+...+1/n -1/(n+1)]

=2[1- 1/(n+1)]

=2n/(n+1)

S10=2×10/(10+1)=20/11