已知F1(x)=2/(1+x),定义Fn+1(x)=F1[Fn(x)],an=[Fn(0)-1]/[Fn(0)+2],则数列an的通项公式是

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/16 10:34:47

已知F1(x)=2/(1+x),定义Fn+1(x)=F1[Fn(x)],an=[Fn(0)-1]/[Fn(0)+2],则数列an的通项公式是
已知F1(x)=2/(1+x),定义Fn+1(x)=F1[Fn(x)],an=[Fn(0)-1]/[Fn(0)+2],则数列an的通项公式是

已知F1(x)=2/(1+x),定义Fn+1(x)=F1[Fn(x)],an=[Fn(0)-1]/[Fn(0)+2],则数列an的通项公式是

F1(0)=2/(1+0)=2
则a1=[2-1]/[2+2]=1/4
Fn(0)=F1[Fn-1(0)]=2/[1+Fn-1(0)]
则:an
=[Fn(0)-1]/[Fn(0)+2]
=[2/(1+Fn-1(0)) -1]/[2/(1+Fn-1(0)) +2]
=[2-(1+Fn-1(0))]/[2+2(1+Fn-1(0))]
=[1-F...

全部展开

F1(0)=2/(1+0)=2
则a1=[2-1]/[2+2]=1/4
Fn(0)=F1[Fn-1(0)]=2/[1+Fn-1(0)]
则:an
=[Fn(0)-1]/[Fn(0)+2]
=[2/(1+Fn-1(0)) -1]/[2/(1+Fn-1(0)) +2]
=[2-(1+Fn-1(0))]/[2+2(1+Fn-1(0))]
=[1-Fn-1(0)]/[4+2Fn-1(0)]
=(-1/2) [Fn-1(0)-1]/[Fn-1(0)+2]
=(-1/2) a(n-1)
故{an}公比是(-1/2)的等比数列
则:an
=a1(-1/2)^(n-1)
=(1/4)*(-1/2)^(n)/[-1/2]
=(-1/2)*(-1/2)^n
=(-1/2)^(n+1)

收起

已知F1(x)=2/(1+x),定义Fn+1(x)=F1[Fn(x)],an=[Fn(0)-1]/[Fn(0)+2],则数列an的通项公式是 已知函数f1(x)=(2x-1)/(x+1) 对于n∈N* 定义fn+1(x)=f1( fn(x)) 求fn(x)解析式苏教版高中数学选修2-2p78页最后一题 设f1(x)=2/(1+x),定义f(n+1)(x)=f1[fn(x)],an=[fn(0)-1]/[fn(0)+2],则a(2007)等于 已知f1(x)=(2x-1)/(x+1),对于n=1,2,…,定义fn+1(x)=f1(fn(x)),若f35(x)=f5(x),则f28(x)=? 已知f(x)=x+1/2 x属于[0,1/2],2(1-x)定义fn(x)=f(fn-1(x))已知f(x)=x+1/2 x属于[0,1/2],2(1-x),x属于【1/2,1],定义fn(x)=f(fn-1(x)),其中f1(x)=f(x),则f2011(1/5)= 已知f1(x)=(2x-1)/(x+1),fn+1(x)=f1[fn(x)](n=1,2,3,……),求f30(x) f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99= 已知f1(x)=e^xsinx,fn(x)= fn-1'(x),n≥2,求f1(0)+f2(0)+f3(0)+ ……f2011(0)的值 已知f1(x)=e^xsinx,fn(x)= fn-1'(x),n≥2,求f1(0)+f2(0)+f3(0)+ ……f2011(0)的值 已知f(x)=x/(x 1),f1(x)=f(x),fn(x)=fn-1[fn-1(x)]求f100(x)的值 f(x)=x/(1+x) x>=0 f1(X)=f(X) fn(X)=fn-1[fn-1(x)]求fn(x)证明:f1(X)+2f2(X)+3f3(x)+……+nfn(X) f1(x)=2/(1+x),f(x)=f1[fn(x)],an=[fn(0)-1]/[fn(0)+2],则a2010=? 定义域和值域均为【0,1】的函数f(x),定义f1(x)=f(x),f2(x)=f(f1(x)),.,fn(x)=f(fn-1(x))n=1,2,3,.满足fn(x)=x的点x【0,1】为f的n段周期点,设f(x)={2x,0 设f1(x)=2/(1+x),fn+1(x)=f1[fn(x)]设f1(x)=2/(1+x),设fn+1(x)=f1〔fn(x)〕,an=〔fn(0)-1〕/〔fn(0)+2〕,n∈N*,求数列{an}的2009项 函数列{fn(x)} 由下列条件定义:f1(x)=根号下x^2+48,fn+1(x)=根号下x^2+6fn(x)(n>=1),求出方程fn(x)=2x的所有实数解 设f(x)=2x+1,f1(x)=f[f(x)],fn(x)=f[fn-1(x)],(n>1,n属于正实数) 求f1(x) f2(x) f3(x)归纳fn(x)表达式 已知n为正整数,规定f1(x)=f(x),fn+1(x)=f(fn(x)),且f(x)=2(1-x),0《x《1;f(x)=x-1,1 已知函数f(x)=(1+x)/(1-3x),f1(x)=f(x),fn+1(x)=f(fn(x)),n大于等于2,n是正整数求f2010(x)