Question

设数列{an}的通项公式为an=2n+1,bn/a1+a2+...an

求数列｛bn｝的前N项和,bn=1/（a1+a2+...an）

Answer 1

bn=1/（2*1+1+2*2+1+...+2*n+1)
=1/[2*(1+2+...+n)+n]
=1/[2*(1+n)*n/2+n]
=1/[n(n+1)+n]
=1/[n(n+2)]
=1/2*[1/n-1/(n+2)]
{bn}的前n项和为：
1/2*[1-1/3+1/2-1/4+1/3-1/5+...+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=1/2*[1+1/2-1/(n+1)-1/(n+2)]
=1/2*[(n+1-1)/(n+1)+(n+2-2)/(2n+4)]
=1/2*[n/(n+1)+n/(2n+4)]
=1/4*n*[2/(n+1)+1/(n+2)]
=1/4*n*(2n+4+n+1)/[(n+1)(n+2)]
=n(3n+5)/[4*(n+1)(n+2)]

Answer 2

a1=3 an等差数列
Sn=a1+a2+...an=(3+2n+1)*n/2=n(n+2)
bn=1/n(n+2)=[1/n-1/(n+2)]/2
数列｛bn｝的前N项和=[1/1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+...+1/n-1/(n+2)]/2
=[1+1/2-1/(n+1)-1/(n+2)]/2

Answer 3

bn=1/[a1+a1+2*1+a1+2*(1+2)……a1+2*(1+2+……n-1)]
bn=1/[a1*n+(n-1)n/2*2]=1/(n*n+2n)=0.5*[1/n-1/(n+2)]

数列｛bn｝的前N项和=0.5*[1.5-1/（n+1)-1/(n+2)]

Answer 4

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