6A - Is the system time invariant?
System: $ Y_k[n]=(k+1)^2\delta[n-(k+1)] $
Time-delay: $ n-n_0 $
Time-delay, then system:
$ T_k[n]=\delta[(n-n_0)-(k+1)] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $
System, then time-delay:
$ T_k[n]=(k+1)^2\delta[n-k] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $
Both $ Y_k $ yield the same output; therefore, the system is time invariant.
6B - What input X[n] would yield the output Y[n]=u[n-1]?
The system detailed above would output $ \delta[n-1] $ given the input of $ \delta[n] $. Therefore, to obtain the output of $ u[n-1] $, the input of $ u[n] $ should be sent through the system. You could also input the definition of $ u[n] $, which is just a sum of many shifted delta functions.
