f(x)=1/x^2=1/[2-(x+2)]^2=1/4 * 1 / [1-(x+2)/2]^2=1/4* {1+2*(x+2)/2 + 3*[(x+2)/2]^2+...+n[(x+2)/2]^(n-1)+. }=1/4+(x+2)/4+3(x+2)^2+...+n(x+2)^(n-1)/ 2^(n+1)+.通项 a_n=n(x+2)^(n-1)/ 2^(n+1),n=1,2,.收敛...