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等差数列{an}.a1=3,d=2,Sn为前n项和,求S=1/s1+1/s2+.+1/sn
题目详情
等差数列{an}.a1=3,d=2,Sn为前n项和,求S=1/s1+1/s2+.+1/sn
▼优质解答
答案和解析
an=3+2（n-1）=2n+1
3+5+……+(2n+1)=(3+2n+1)*n/2=n(n+2)
所以1/[3+5+……+(2n+1)]=1/n(n+2)=(1/2)*[1/n-1/(n+2)]
S=1/s1+1/s2+.+1/sn
=1/2*[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+……+1/(n-1) - 1/(n+1)+1/n - 1/(n+2)]
=1/2*[1+1/2- 1/(n+1)- 1/(n+2)]
=3/4-(2n+3)/[2(n+1)(n+2)]
=.
主要是思路 ,运算的问题可以自己解决吧.
3+5+……+(2n+1)=(3+2n+1)*n/2=n(n+2)
所以1/[3+5+……+(2n+1)]=1/n(n+2)=(1/2)*[1/n-1/(n+2)]
S=1/s1+1/s2+.+1/sn
=1/2*[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+……+1/(n-1) - 1/(n+1)+1/n - 1/(n+2)]
=1/2*[1+1/2- 1/(n+1)- 1/(n+2)]
=3/4-(2n+3)/[2(n+1)(n+2)]
=.
主要是思路 ,运算的问题可以自己解决吧.
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