将f(x)=x/(x^2-2x-3)展开①f(x)=x/(x^2-2x-3)=1/4[3/(x-3)+1/(x+1)]=(-1/4)*[1/(1-x/3)]+(1/4)*[1/(1+x)]=(-1/4)∑(n=0)(x/3)^n+(1/4)∑(n=0)(-x)^n②f(x)=x/(x^2-2x-3)=(-x/12)*[1/(1-x/3)]-(x/4)[1/(1+x)]=(-1/4)∑(n=0)(x/3)^(n+1)+(1/4)∑(n=0)(-x)^(n+1)
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/16 05:56:01
将f(x)=x/(x^2-2x-3)展开①f(x)=x/(x^2-2x-3)=1/4[3/(x-3)+1/(x+1)]=(-1/4)*[1/(1-x/3)]+(1/4)*[1/(1+x)]=(-1/4)∑(n=0)(x/3)^n+(1/4)∑(n=0)(-x)^n②f(x)=x/(x^2-2x-3)=(-x/12)*[1/(1-x/3)]-(x/4)[1/(1+x)]=(-1/4)∑(n=0)(x/3)^(n+1)+(1/4)∑(n=0)(-x)^(n+1)
将f(x)=x/(x^2-2x-3)展开
①f(x)=x/(x^2-2x-3)=1/4[3/(x-3)+1/(x+1)]=(-1/4)*[1/(1-x/3)]+(1/4)*[1/(1+x)]=(-1/4)∑(n=0)(x/3)^n+(1/4)∑(n=0)(-x)^n
②f(x)=x/(x^2-2x-3)=(-x/12)*[1/(1-x/3)]-(x/4)[1/(1+x)]=(-1/4)∑(n=0)(x/3)^(n+1)+(1/4)∑(n=0)(-x)^(n+1)
为什么有2个答案啊 哪里出错了
将f(x)=x/(x^2-2x-3)展开①f(x)=x/(x^2-2x-3)=1/4[3/(x-3)+1/(x+1)]=(-1/4)*[1/(1-x/3)]+(1/4)*[1/(1+x)]=(-1/4)∑(n=0)(x/3)^n+(1/4)∑(n=0)(-x)^n②f(x)=x/(x^2-2x-3)=(-x/12)*[1/(1-x/3)]-(x/4)[1/(1+x)]=(-1/4)∑(n=0)(x/3)^(n+1)+(1/4)∑(n=0)(-x)^(n+1)
①的首项为0,去掉首项即为
(-1/4)∑(n=1)(x/3)^n+(1/4)∑(n=1)(-x)^n.
②就是
(-1/4)∑(n=1)(x/3)^n+(1/4)∑(n=1)(-x)^n.
两者是一样的.
，第二个n初值为-1，或第一个n初值为1，这样就是一样了，只是形式不同罢了。
x/[(x-3)(x+1)]