函数f(x)=1/x^2+3x+2展开为x+3的幂级数

为方便,记t=x+3
f(x)=1/[(x+1)(x+2)]
=1/(x+1)-1/(x+2)
=1/(x+3-2)-1/(x+3-1)
=1/(t-2)+1/(1-t)
=-0.5/(1-t/2)+1/(1-t)
=-0.5[1+t/2+(t/2)^2+...+(t/2)^n+...]+[1+t+t^2+...t^n+...]
=0.5+t(1-1/4)+t^2(1-1/8)+.+t^n(1-1/2^(n+1))+...
=0.5+t*3/4+t^2*7/8+.+t^n*[2^(n+1)-1]/2^(n+1)+...