作业帮f1(x)=2/(1+x),f(x)=f1[fn(x)],an=[fn(0)-1]/[fn(0)+2],则a2010=?
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/12 03:59:48
![f1(x)=2/(1+x),f(x)=f1[fn(x)],an=[fn(0)-1]/[fn(0)+2],则a2010=?](/uploads/image/z/10886727-39-7.jpg?t=f1%28x%29%3D2%2F%281%2Bx%29%2Cf%28x%29%3Df1%5Bfn%28x%29%5D%2Can%3D%5Bfn%280%29-1%5D%2F%5Bfn%280%29%2B2%5D%2C%E5%88%99a2010%3D%3F)
f1(x)=2/(1+x),f(x)=f1[fn(x)],an=[fn(0)-1]/[fn(0)+2],则a2010=?
f1(x)=2/(1+x),f(x)=f1[fn(x)],an=[fn(0)-1]/[fn(0)+2],则a2010=?
f1(x)=2/(1+x),f(x)=f1[fn(x)],an=[fn(0)-1]/[fn(0)+2],则a2010=?
f1(x) = 1/(1+x),a1 = 1/4
f2(x)= 2(1+x) / (3+x) a2 = -1/8
f3(x) = 2(3+x)/(5+3x),a3 = 1/16
f4(x)=2(5+3x)/(11+5x),a4 = -1/32
an = (-1)^(n+1) / 2^(n+1)
a2010= -1/2^2011