f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99=
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 04:26:28
![f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99=](/uploads/image/z/6894607-31-7.jpg?t=f1%28x%29%3D2%2F%28x%2B1%29%2C%E8%80%8Cfn%2B1%3Df1%5Bfn%28x%29%5D%2C%E8%AE%BEan%3D%5Bfn%282%29-1%5D%2F%5Bfn%282%29%2B2%5D%2C%E5%88%99a99%3D)
xO;N@J]v(^_$ZKb˕.R$;
:
J$T\!^
fuV
yCσ6m38LF6~:bWp,&/"7l%#(HB&|K^,
yB0f|XPm=ퟆ`AW
f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99=
f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99=
f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99=
fn+1=2/[fn(x)+1] f1(2)=2/3 a1=-1/8,f2(2)=6/5,a2=1/16,f3(2)=10/11,a3=-1/32,所以通式an=(-1)^n*1/[2^(n+2)];a99=-1/2^101;(说明:2^n,2的n次方