等差数列{an}.a1=3,d=2,Sn为前n项和,求S=1/s1+1/s2+.+1/sn_数学解答

等差数列{an}.a1=3,d=2,Sn为前n项和,求S=1/s1+1/s2+.+1/sn

人气：202 ℃　时间：2020-10-02 01:20:53

解答

Sn=n(3+2n+1)/2 (求和公式,a1=3 an=2n+1)
Sn=n^2+2n1/Sn=1/(n^2+2n)
1/(n^2+2n)=1/n(n+2)=1/2*{1/n-1-1/(n+2)}
Sn=1/2{1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.+1/n-1/(n+2)}
=3/4-(2n+3)/2(n+1)(n+2)