Question

已知数列{an}的前n项和为Sn=1/2n2+1/2n(n？n*)(1)求数列{an}的通项公式。(2)记T=1/s1+1/s2+1/s3+…+1/s...

已知数列{an}的前n项和为Sn=1/2n2+1/2n(n？n*)(1)求数列{an}的通项公式。(2)记T=1/s1+1/s2+1/s3+…+1/s99.求T的值

Answer 1

an=Sn-Sn-1=[1/2n2+1/2n]-[1/2(n-1)^2+1/2(n-1)]=n an=n
Sn=1/2n2+1/2n 1/Sn=2/n(n+1)=2[1/n-1/n+1]
(2)记T=1/s1+1/s2+1/s3+…+1/s99=2[1-1/2+1/2-/13+……+1/99-1/100]=2*(1-1/100)=99/50

Answer 2

Sn-S(n-1)=(1/2n^2+1/2n)-(1/2(n-1)^2+1/2(n-1))=An(n>=2)
An=n(n>=2)
A1=s1=1 ,符合
得an=n
Sn=(1+n)n/2
T=2/1*2+2/2*3+...+2/99*100=2[1-1/2+1/2-1/3+...+1/99-1/100]=2[1-1/100]=99/50

Answer 3

2