Question

等差数列an中，a1＝3/2.d＝1.前n项和为Sn.求1/s1+1/s2+...+1/Sn

求和用裂项相消除法

Answer 1

解答：
等差数列an中，a1＝3/2.d＝1
∴ Sn=na1+n(n-1)*d/2
=(3/2)n+(n²-n)/2
=(n²+2n)/2
∴ 1/Sn=2/(n²+2n)=2/[n(n+2)]=1/n-1/(n+2)
∴ 1/s1+1/s2+...+1/Sn
=1/1-1/3+1/2-1/4+1/3-1/5+.....+1/(n-1)-1/(n+1)+1/n-1/(n+2)
=1+1/2-1/(n+1)-1/(n+2)
=3/2-1/(n+1)-1/(n+2)

Answer 2

a1=3/2
d=1
Sn=1/2(n+2)n
1/Sn=2[1/(n+2)n]=2[1/2(1/(n+2))-1/n]
1/S(n-1)=2*1/2[(1/(n+1)-1/(n-1)]
1/S(n-2)=2*1/2[(1/n - 1/(n-2)]
1/S(n-3)=2*1/2[1/(n-1)-1/(n-3)]
.... = 2*1/2[.......-........]
1/s4=2*1/2(1/6-1/4)
1/S3=2*1/2(1/5-1/3)
1/S2=2*1/2[1/4-1/2]
1/S1=2*1/2(1/a3-1/1)
规律就是这个右边加起来得到
2*1/2(1/(n+2)+1/(n+1)-1/2-1)
化简得到
(2n+3)/(n^2+3n+1) -3/2

Answer 3

由题意可算出Sn=n(n+2)/2
1/s1+1/s2+...+1/Sn
=1/2(2/n-2/（n+2）+ .....1)
=1/2(3-1/(n+1)-2/(n+1))