利用已知级数 1/(1-x) = ∑(n≥1)[x^(n-1)],|x|<1,可得 f(x) = 1/(x²+4x+3) = (1/2)[1/(x+1)-1/(x+3)] = (1/2)[1/(x+1)]-(1/4){1/[1+(x+1)/2]} = (1/2)[1/(x+1)]-(1/4)∑(n≥1){[-(x+1)/2]^(n-1)},0<|x+1|<1。