若数列an满足a(n+1)≤f(an)(n为正整数),数列bn满足bn=an/2n+1.是证明b+b2+.+bn≤1/2
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/22 20:59:10
若数列an满足a(n+1)≤f(an)(n为正整数),数列bn满足bn=an/2n+1.是证明b+b2+.+bn≤1/2若数列an满足a(n+1)≤f(an)(n为正整数),数列bn满足bn=an/2
若数列an满足a(n+1)≤f(an)(n为正整数),数列bn满足bn=an/2n+1.是证明b+b2+.+bn≤1/2
若数列an满足a(n+1)≤f(an)(n为正整数),数列bn满足bn=an/2n+1.是证明b+b2+.+bn≤1/2
若数列an满足a(n+1)≤f(an)(n为正整数),数列bn满足bn=an/2n+1.是证明b+b2+.+bn≤1/2
这个高数难了~~~~
若数列an满足a(n+1)≤f(an)(n为正整数),数列bn满足bn=an/2n+1.是证明b+b2+.+bn≤1/2
已知函数f(x)=x/(x+1),若数列{An}(n属於正整数)满足A1=1,A(n+1)=f(An)(1)设bn=1/An,求证数列{bn}是等差数列,(2)求数列{An}的通向公式An(3)设数列{Cn}满足:Cn=2^n/An,求数列{C
已知函数f(x)=x/x+3,数列an满足a1=1,a(n+1)=f(an) (n属于N+)已知函数f(x)=x/x+3,数列an满足a1=1,a(n+1)=f(an) (n属于N+)求 1)数列{an}的通项公式2)若数列{bn}满足bn=(1/2)an*a(n+1)*3^n,Sn=b1+b2+b3+...+bn
已知函数f(x)=x/x+3,数列an满足a1=1,a(n+1)=f(an) (n属于N+)已知函数f(x)=x/x+3,数列an满足a1=1,a(n+1)=f(an) (n属于N+)求 1)数列{an}的通项公式2)若数列{bn}满足bn=(1/2)an*a(n+1)*3^n,Sn=b1+b2+b3+...+bn
数列[An]满足a1=2,a(n+1)=3an-2 求an
已知函数f(x)=3x+2,数列{an}满足:a1不等于-1且an+1=f(an)(n属于正整数),若数列{an+c}是等比数列...已知函数f(x)=3x+2,数列{an}满足:a1不等于-1且an+1=f(an)(n属于正整数),若数列{an+c}是等比数
若数列an满足a1=1,an+1=2^nan...若数列an满足a1=1,a(n+1)=2^n·an,则数列an的通项公式?
设数列{an}满足a(n+1)=2an+n^2-4n+1若a1=3,求证:存在f(n)=an^2+bn+c(a,b,c为常数),使数列{an+f(n)}是等比数列,并求出数列{an}的通项公式
已知数列an满足条件a1=-2 a(n+1)=2an/(1-an) 则an=
已知数列{an}满足a1=33,a(n+1)-an=2n,求an/n的最小值
已知数列an满足a1=2,an=a(n-1)+2n,(n≥2),求an
数列{an},定义数列满足:Δan=a(n+1)-an,定义数列{(Δan)的平方}满足:(Δan)的平方=Δa(n+1)-Δan,若数列{2^Δan}中各项均为1,且a21=a2012=0,则a1=?若数列{(Δan)的平方}中各项均为1 不好意思,
对于数列{an},定义{Δan}为数列{an}一阶差分数列,其中Δan=a(n+1)-an若数列{an}的首项是1,且满足Δan-an=2^n,证明数列{an/2^n}为等差数列
已知数列{an}满足a1=2,a(n+1)-an=a(n+1)*an,则a31=?
已知数列{An}满足a1=1,a(n+1)=2an+1 求证数列{an+1}是等比数列 求数列{an}通式
定义:若数列{An}满足An+1=An2,则称数列{An}为“平方递推数列”.已知数列{an}中,a1=2,点(an,an+1)在函数f(x)=2x2+2x的图象上,其中n为正整数.(Ⅰ)证明:数列{2an+1}是“平方递推数列”,且数列{lg
在数列an中,a1=1,且满足a(n+1)=3an +2n,求an
数列﹛an﹜满足an=2a(n+1)-1,怎么证明等差?