等差数列题目在等差{an}中an=3n 1,求{1\(an*an 1}前n项和Sn.

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等差数列题目
在等差{an}中an=3n 1,求{1\(an*an 1}前n项和Sn.

漏符号了,如果是an=3n+1,则
1/[ana(n+1)]=1/[(3n+1)(3n+4)]=(1/3)[1/(3n+1)-1/(3n+4)]
Sn=1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]
=(1/3)[1/4-1/7+1/7-1/10+...+1/(3n+1)-1/(3n+4)]
=(1/3)[1/4 -1/(3n+4)]
=1/12 -1/[3×(3n+4)]
如果是an=3n-1,则
1/[ana(n+1)]=1/[(3n-1)(3n+2)]=(1/3)[1/(3n-1)-1/(3n+2)]
Sn=1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]
=(1/3)[1/2-1/5+1/5-1/8+...+1/(3n-1)-1/(3n+2)]
=(1/3)[1/2 -1/(3n+2)]
=1/6 -1/[3×(3n+2)]

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