1/(k^4-10k^2+9)
=1/{(k^2-9)(k^2-1)}
=(1/8){1/(k^2-9)-1/(k^2-1)}
=(1/8){1/(k-3)(k+3)}-(1/8){1/(k-1)(k+1)}
=(1/48){1/(k-3)-1/(k+3)}-(1/16){1/(k-1)-1/(k+1)}

を利用して
与式=(1/48){1+1/2+1/3+1/4+1/5+1/6-1/(n-1)-1/n-1/n-1/(n+1)-1/(n+2)-1/(n+3)}-(1/16){1/3+1/4-1/n-1/(n+1)}
=(1/48)(1+1/2-2/3-2/4+1/5+1/6)-(1/48){1/(n-2)+1/(n-1)-2/n-2/(n+1)+1/(n+2)+1/(n+3)}
=(1/48){7/10-1/(n-2)-1/(n-1)+2/n+2/(n+1)-1/(n+2)-1/(n+3)}