Question

函数f(x)=1/(x2-x-6)展开成X的幂级数

Answer 1

f(x)=1/(x-3)(x+2)=(1/5){[1/(x-3)]-[1/(x+2)]}=(1/5){[(-1/3)/(1-x/3)]-[(1/2)/(1+x/2)]}

=(1/5){(-1/3)∑(x/3)^n-(1/2)∑(-x/2)^n}=-1/15∑[(1/3)^n-(-1/2)^n]x^n

(-2