f(x)=1/(x^2-3x-4)=1/[(x-4)(x+1)] = (1/5)[1/(x-4)-1/(x+1)]= (-1/5)[1/(4-x)+1/(1+x)] = (-1/5){1/[3-(x-1)]+1/[2+(x-1)]}= (-1/5){(1/3)/[1-(x-1)/3]+(1/2)/[1+(x-1)/2]}= (-1/5)∑ (1/3)[(x-1)/3]^n+(-1)^n(1/2)[(x-1)/2]^n= (-1/5)∑ [1/3^(n+1)+(-1)^n/2^(n+1)])(x-1)^n
