Linearity and Time Invariance
Input Output Xk[n]=d[n-k] Yk[n]=(k+1)^2 d[n-(k+1)]
Part A
Since the System squares the k term this would make the system time variant. xk[n-k]->system->Yk[n]=(k+1)^2 d[n-(k+1)]->time shift->Yk[n]=(k+1)^2 d[n-(k+1)+to] xk[n-k+to]->system->Yk[n]=(k+to+1)^2 d[n-(k+1)+to] the two expressions are not equal above, so the system is time invariant.
Part B
The input of the system would be u[n].
