
大学高数题
3个回答
展开全部
f(x)=1/(x²-3x+2)
=1/[(x-1)(x-2)]
=1/(x-2)-1/(x-1)
=1/(1-x)-1/(2-x)
=1/(1-x)-(1/2)/(1-x/2)
=Σx^n-(1/2)Σ(x/2)^n n=0→∞
=Σ[1-1/2^(n+1)]x^n n=0→∞
=1/[(x-1)(x-2)]
=1/(x-2)-1/(x-1)
=1/(1-x)-1/(2-x)
=1/(1-x)-(1/2)/(1-x/2)
=Σx^n-(1/2)Σ(x/2)^n n=0→∞
=Σ[1-1/2^(n+1)]x^n n=0→∞
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询