The Question is given like this Find the inverse Z transform of
$$X(z)=\frac{z^{2}}{(z-1)^{3}}$$
Attempt: $$\frac{X(z)}{z}=\frac{z}{(z-1)^{3}}=\frac{A}{(z-1)}+\frac{B}{(z-1){2}}+\frac{c}{(z-1){3}}$$ after solving A=0,B=1,C=1 $$\frac{X(z)}{z}=\frac{1}{(z-1){2}}+\frac{1}{(z-1){3}}$$ $$X(z)=\frac{z^{-1}}{(1-z^{-1})^{2}}+\frac{z^{-2}}{(1-z^{-1})^{3}}$$ Then taking its inverse Z transform we get $$X(n)=nu(n)+\frac{(n-1)^{2}}{{2}}u(n-1)$$ BUT
the exact ans i am getting from all different method is
$$\frac{n(n+1)}{2}$$
Whats is the mistake i am doing in this partial fraction method?
PS: I got the ans of this question from all possible method like convlution method,residue method,etc so plz help me with only partial fraction method as i already know other method.