If $\frac{(1+x)(1+2x)(1+3x)}{(1-x)(1-5x+6x^2)}=k+\frac{A}{1-x}+\frac{B}{1-2x}+\frac{C}{1-3x}$ then

$(1+x)(1+2x)(1+3x)=k(1-x)(1-2x)(1-3x)$

$+A(1-2x)(1-3x)+B(1-x)(1-3x)+C(1-x)(1-2x)$

Compare coef. of $x^3, x^2$..... etc .....

Substitute $x=\frac{1}{3}$

$\left(\frac{4}{3}\right)\left(\frac{5}{3}\right)\left(2\right)=C\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)\Rightarrow c=20$

Comparing the coefficient of $x^3$ 

we get $6=-6k\Rightarrow k=-1$

It implies that $c+k=20-1=19$