设数列{an}满足an+1/an=n+2/n+1,且a1=21.求数列an的通项公式.2.设bn=an/2^n,数列{bn}的前n项和为.
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/19 21:13:22
![设数列{an}满足an+1/an=n+2/n+1,且a1=21.求数列an的通项公式.2.设bn=an/2^n,数列{bn}的前n项和为.](/uploads/image/z/4516244-44-4.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3an%2B1%2Fan%3Dn%2B2%2Fn%2B1%2C%E4%B8%94a1%EF%BC%9D21%EF%BC%8E%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.2%EF%BC%8E%E8%AE%BEbn%EF%BC%9Dan%EF%BC%8F2%EF%BC%BEn%2C%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BA.)
设数列{an}满足an+1/an=n+2/n+1,且a1=21.求数列an的通项公式.2.设bn=an/2^n,数列{bn}的前n项和为.
设数列{an}满足an+1/an=n+2/n+1,且a1=2
1.求数列an的通项公式.2.设bn=an/2^n,数列{bn}的前n项和为.
设数列{an}满足an+1/an=n+2/n+1,且a1=21.求数列an的通项公式.2.设bn=an/2^n,数列{bn}的前n项和为.
1、a(n+1)/an=(n+2)/(n+1)
a(n+1)/(n+2)=an/(n+1)
设cn=an/(n+1) 则c(n+1)=a(n+1)/(n+2),且c1=a1/(1+1)=1
即c(n+1)=cn=.=c1=1
故an/(n+1)=1
an=n+1
2、因bn=an/2^n=(n+1)/2^n
设数列{bn}的前n项和为Tn
Tn=b1+b2+.+bn
=2/2+3/2^2+.+(n+1)/2^n
2Tn=2/1+3/2+4/2^2+...+(n+1)/2^(n-1)
后式减前式:Tn=2+1/2+1/2^2+.+1/2^(n-1)-(n+1)/2^n
=2+(1/2)[1-(1/2)^(n-1)]/(1-1/2)-(n+1)/2^n
=3-1/2^(n-1)-(n+1)/2^n
=3-(n+3)/2^n
1、由已知得an/an-1=(n+1)/n,an-1/an-2=n/(n-1)……a2/a1=3/2。
将以上各式左右分别全部相乘,得an/a1=(n+1)/2,由a1=2,得an=n+1
2、Sn为bn前n项和,Sn=(1/2)^n *(n+1)+……+(1/2)*(1+1),Sn*(1/2)=(1/2)^(n+1) *(n+1)+(1/2)^n *n,得Sn-Sn*(1...
全部展开
1、由已知得an/an-1=(n+1)/n,an-1/an-2=n/(n-1)……a2/a1=3/2。
将以上各式左右分别全部相乘,得an/a1=(n+1)/2,由a1=2,得an=n+1
2、Sn为bn前n项和,Sn=(1/2)^n *(n+1)+……+(1/2)*(1+1),Sn*(1/2)=(1/2)^(n+1) *(n+1)+(1/2)^n *n,得Sn-Sn*(1/2)=-(1/2)^(n+1) *(n+1)+(1/2)^n+(1/2)^(n-1)+……+(1/2)),
得(1/2)*Sn=(1/2)-(1/2)^(n+1) *(n+1)+(1-(1/2)^n)/4,可得出Sn。
注:计算过程比较粗略,未经验算不一定准确,计算的思路是一定的,即求数列cn=an*bn之和Sn,其中an为等差数列,bn为等比数列,均可用Sn-q*Sn的化简方法,略去中间不可求和项,实现求和的终级目标,呵呵~~~
收起
1)a(n+1)/(n+2)=a(n)/(n+1)
设b(n)=a(n)/(n+1) 则b(n+1)=a(n+1)/(n+2)
易知b(n+1)-b(n)=0
b(n)=1,所以a(n)=n+1
2)Sn=b1+b2+....+bn
=a1/2+a2/4+....+an/2^n
用错位相减法:结果=3-(n+3)/2^n