1/(x²-5x+6)
=1/[(x-2)(x-3)]
=1/(x-3)-1/(x-2)
=1/(2-x)-1/(3-x)
=1/(1-(x-1))-1/(2-(x-1))
=1/(1-(x-1))-(1/2)/(1-(x-1)/2)
=Σ(x-1)^n-(1/2)Σ[(x-1)/2]^n n=0到+∞
=Σ(x-1)^n-Σ(1/2^(n+1))(x-1)^n n=0到+∞
=Σ(1-1/2^(n+1))(x-1)^n n=0到+∞
将f(X)＝1/x²-5X+6展成（X＝1）的幂级数
1/(x²-5x+6)
=1/[(x-2)(x-3)]
=1/(x-3)-1/(x-2)
=1/(2-x)-1/(3-x)
=1/(1-(x-1))-1/(2-(x-1))
=1/(1-(x-1))-(1/2)/(1-(x-1)/2)
=Σ(x-1)^n-(1/2)Σ[(x-1)/2]^n n=0到+∞
=Σ(x-1)^n-Σ(1/2^(n+1))(x-1)^n n=0到+∞
=Σ(1-1/2^(n+1))(x-1)^n n=0到+∞