Question

已知数列a(n)是等差数列,且a1=2,a1+a2+a3=12.求1/s1+1/s2+1/s3+...+1/s10的值

Answer 1

a1=2,a1+a2+a3=3a1+3d=12

d=2

sn=na1+(n-1)nd/2=n^2+n=n(n+1)

1/s1+1/s2+1/s3+...+1/s10

=1/(1x2)+1/(2x3)+1/(3x4)+.1/(10x11)

=1-1/2+1/2-1/3+1/3-1/4.+1/10-1/11

=1-1/11

=10/11