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.
在等差{an}中an=3n 1,求{1\(an*an 1}前n项和Sn.
![等差数列题目在等差{an}中an=3n 1,求{1\(an*an 1}前n项和Sn.](/uploads/image/z/18643449-57-9.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E9%A2%98%E7%9B%AE%E5%9C%A8%E7%AD%89%E5%B7%AE%7Ban%7D%E4%B8%ADan%3D3n+1%2C%E6%B1%82%7B1%5C%28an%2Aan+1%7D%E5%89%8Dn%E9%A1%B9%E5%92%8CSn.)
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)]
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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