0已知满足a1=1,a2=2,an+2=(an+1+an)/2 (1)令bn=an+1-an,证明是等比数列 (2)求an的通项公式.

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/18 19:44:16
0已知满足a1=1,a2=2,an+2=(an+1+an)/2(1)令bn=an+1-an,证明是等比数列(2)求an的通项公式.0已知满足a1=1,a2=2,an+2=(an+1+an)/2(1)令

0已知满足a1=1,a2=2,an+2=(an+1+an)/2 (1)令bn=an+1-an,证明是等比数列 (2)求an的通项公式.
0已知满足a1=1,a2=2,an+2=(an+1+an)/2 (1)令bn=an+1-an,证明是等比数列 (2)求an的通项公式.

0已知满足a1=1,a2=2,an+2=(an+1+an)/2 (1)令bn=an+1-an,证明是等比数列 (2)求an的通项公式.
你把an+2=(an+1+an)/2变形一下,就可以得出bn+1/bn =(an+2 - an+1)/(an+1 - an)= -1/2,然后看b1 = a2 - a1 =1不为0,所以bn不就是首项为1,公比为-1/2的等比数列么?
bn = (-1/2)^(n-1) (表示(-1/2)的(n-1)次方),然后
a2 - a1 = (-1/2)^(1-1)
a3 - a2 =(-1/2)^(2-1)

an - an-1 = (-1/2)^(n-1-1)
然后相加,不就是an - a1 = (2/3)[1 - (-1/2)^(n-1)],an不就算出来了么?

1,∵a(n+2)=[a(n+1)+an]/2
∴2a(n+2)=a(n+1)+an
∴2a(n+2)-2a(n+1)=-a(n+1)+an
即2[a(n+2)-a(n+1)]=-[a(n+1)-an]
也即2b(n+1)=-bn
∴b(n+1)/bn=-1/2,为常数
而b1=a2-...

全部展开

1,∵a(n+2)=[a(n+1)+an]/2
∴2a(n+2)=a(n+1)+an
∴2a(n+2)-2a(n+1)=-a(n+1)+an
即2[a(n+2)-a(n+1)]=-[a(n+1)-an]
也即2b(n+1)=-bn
∴b(n+1)/bn=-1/2,为常数
而b1=a2-a1=2-1=1
∴bn是以1为首项、-1/2为公比的等比数列
2,bn=(-1/2)^(n-1)=a(n+1)-an
∴an-a(n-1)=(-1/2)^(n-2)
a(n-1)-a(n-2)=(-1/2)^(n-3)
………………………………
a2-a1=(-1/2)^0
累加,得:an-a1=(-1/2)^(n-2)+(-1/2)^(n-3)+……+1
=[1-(-1/2)^(n-1)]/(1+1/2)
=2/3-2/3*(-1/2)^(n-1)
所以an=5/3-2/3*(-1/2)^(n-1)

收起

已知数列满足a1=1/2,an+1=2an/(an+1),求a1,a2已知数列满足a1=1/2,a(n+1)=2an/(an+1),求a1,a2;证明0 已知数列an满足an=1+2+...+n,且1/a1+1/a2+...+1/an 已知数列an满足a1=0 a2=1 an=(An-1+An-2)/2 求liman 已知数列an满足a1=0 a2=1 an=(An-1+An-2)/2 求liman 数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an 已知数列an'满足a1=1/2,a1+a2+a3+...+an=n^2an,求通项公式 已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an求an 已知正数a1,a2,a3...an满足a1*a2*a3*...*an=1.求证:(2+a1)*(2+a2)*(2+a3)*...*(2+an)>=3^n 已知数列{an}满足a1=2且anan+1-2an=0球a2,a3,a4的值 已知a1+a2+…….+an=1求证:a1^2/(a1+a2) + a2^2/(a2+a3)…….+an-1^2/(an-1+an) +an^2/(an+a1)>1/2已知a1+a2+…….+an=1求证:a1^2/(a1+a2) + a2^2/(a2+a3)……+an-1^2/(an-1+an) +an^2/(an+a1)>1/2 已知数列{an}中满足a1=1,a(n+1)=2an+1 (n∈N*),证明a1/a2+a2/a3+…+an/a(n+1) 已知数列{an}满足a1=0,a2=1,an+2=3an+1-2an,则{an}的前n项和Sn=( ) 关于数列极限的已知数列an满足a1=0 a2=1 an=(an-1+an-2)/2 求lim(n->无穷)an 已知等差数列(an)满足a1=2,且a1,a2,a5成等比数列 已知递增数列{an}满足a1=1,(2an+1)=an+(an+2),且a1,a2,a4成等比数列.求an 已知数列an满足an=1+2+...n,且(1/a1)+(1/a2)+...(1/an) 等比数列an满足 lim(a1+a2+a3+...+an)=1/2 求a1取值范围 1.用数学归纳法证明:(a1+a2+a3+.+an)^2=a1^2+a2^2+.+an^2+2(a1*a2+a1*a3+.+an-1*an)2.已知数列{an}满足a1=0.5,a1+a2+a3+.+an=Sn=n^2*an(n属于N*),试用数学归纳法证明an=1/(n(n+1))