Correct choice is (c) z^-k \([X^+(z)+\sum_{n=1}^k x(-n)z^n]\); k>0
The explanation is: From the definition of one sided z-transform we have,
Z^+{x(n-k)}=\(z^{-k}[\sum_{l=-k}^{-1} x(l) z^{-l}+\sum_{l=0}^{\infty} x(l)z^{-l}]\)
=\(z^{-k}[\sum_{l=-1}^{-k} x(l) z^{-l}+X^+ (z)]\)
By changing the index from l to n= -l, we obtain
Z^+{x(n-k)}=\(z^{-k}[X^+(z)+\sum_{n=1}^k x(-n)z^n]\) ;k>0