神马作文网高中数列求和的问题
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高中数列求和的问题
a(n+1)=an/(an +1)
1/a(n+1)=(an +1)/an=1/an +1
1/a(n+1)-1/an=1,为定值
1/a1=1/1=1,数列{1/an}是以1为首项,1为公差的等差数列
1/an=1+1×(n-1)=n
bn=1/an,数列{bn}是以1为首项,1为公差的等差数列.
an=1/n
cn=an/(n+1)=(1/n)/(n+1)=1/[n(n+1)]=1/n -1/(n+1)
Sn=c1+c2+...+cn
=1/1-1/2+1/2-1/3+...+1/n -1/(n+1)
=1- 1/(n+1)
=n/(n+1)
n=1时,a1=S1=1²+2×1+1=4
n≥2时,an=Sn-S(n-1)=(n²+2n+1)-[(n-1)²+2(n-1)+1]=2n+1
n=1时,a1=2×1+1=3≠4
数列{an}的通项公式为
an=4 n=1
2n+1 n≥2
n=1时,T1=1/(a1·a2)=1/[4·(2·2+1)]=1/20
n≥2时,1/[an·a(n+1)]=1/[(2n+1)(2(n+1)+1)]=(1/2)[1/(2n+1)-1/(2(n+1)+1)]
Tn=T1+ 1/(a2·a3)+1/(a3·a4)+...+1/[an·a(n+1)]
=1/20 +(1/2)[1/(2×2+1)-1/(2×3+1)+1/(2×3+1)-1/(2×4+1)+...+1/(2n+1)-1/(2(n+1)+1)]
=1/20+(1/2)[1/5 -1/(2n+3)]
=3/20 -1/[2(2n+3)]
n=1时,T1=3/20 -1/[2(2×1+3)]=3/20 -1/10=1/20,同样满足
综上,得Tn=3/20 -1/[2(2n+3)]
1/a(n+1)=(an +1)/an=1/an +1
1/a(n+1)-1/an=1,为定值
1/a1=1/1=1,数列{1/an}是以1为首项,1为公差的等差数列
1/an=1+1×(n-1)=n
bn=1/an,数列{bn}是以1为首项,1为公差的等差数列.
an=1/n
cn=an/(n+1)=(1/n)/(n+1)=1/[n(n+1)]=1/n -1/(n+1)
Sn=c1+c2+...+cn
=1/1-1/2+1/2-1/3+...+1/n -1/(n+1)
=1- 1/(n+1)
=n/(n+1)
n=1时,a1=S1=1²+2×1+1=4
n≥2时,an=Sn-S(n-1)=(n²+2n+1)-[(n-1)²+2(n-1)+1]=2n+1
n=1时,a1=2×1+1=3≠4
数列{an}的通项公式为
an=4 n=1
2n+1 n≥2
n=1时,T1=1/(a1·a2)=1/[4·(2·2+1)]=1/20
n≥2时,1/[an·a(n+1)]=1/[(2n+1)(2(n+1)+1)]=(1/2)[1/(2n+1)-1/(2(n+1)+1)]
Tn=T1+ 1/(a2·a3)+1/(a3·a4)+...+1/[an·a(n+1)]
=1/20 +(1/2)[1/(2×2+1)-1/(2×3+1)+1/(2×3+1)-1/(2×4+1)+...+1/(2n+1)-1/(2(n+1)+1)]
=1/20+(1/2)[1/5 -1/(2n+3)]
=3/20 -1/[2(2n+3)]
n=1时,T1=3/20 -1/[2(2×1+3)]=3/20 -1/10=1/20,同样满足
综上,得Tn=3/20 -1/[2(2n+3)]