在等差数列中,a1=3,d=2,Sk为前k项和,则S=1/s1+1/s2+1/s3.1/sn=_______数学解答

在等差数列中,a1=3,d=2,Sk为前k项和,则S=1/s1+1/s2+1/s3.1/sn=______

人气：178 ℃　时间：2020-04-16 05:22:24

解答

Sn=3n+2n(n-1)/2=3n+n²-n=n²+2n=n(n+2)
1/Sn=1/[n(n+2)]=(1/2)[1/n-1/(n+2)]
S=1/s1+1/s2+1/s3+...+1/sn
=1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/(n-2)-1/n+1/(n-1)-1/(n+1)+1/n-1/(n+2)
=1+1/2-1/(n+1)-1/(n+2)
=n(3n+5)/[2(n+1)(n+2)]