Linearity and Time invariant system

Time invariance

The system shown is not time invariance system. Although there is no $ t\, $ variable as the coefficient of the output or the system, the system depends on the delay of the input.

The input is shifted before undergoing the transformation

$ X[n] = X_{k + \alpha} = \delta[n - k - \alpha] = \delta[n - (k + \alpha)] \, $

$ Y[n] = Y_{k + \alpha} = (k + \alpha + 1)^2\delta[n - (k + \alpha + 1)] \, $

If the signal is shifted at the output, $ Y[n] = Y_k[n - \alpha] = (k+1)^2\delta[n-(k+1) - \alpha] = (k+1)^2\delta[n-(k + 1 + \alpha)]\, $

Both signal doesn't match, as the coeefeicent change according to the time when the signal is inputted. Therefore, the system is not time invariant

Finding the input

$ u[n-1] = \sum^\inf_{k=1} \delta[n-k] $

$ = \sum^\inf_{k=0} \delta[n-(k+1)] $

$ = \sum^\inf_{k=0} \frac{(k+1)^2}{(k+1)^2}\delta[n-(k+1)] $

$ = \frac{1}{(k+1)^2}\sum^\inf_{k=0} (k+1)^2\delta[n-(k+1)] $

$ = \frac{1}{(k+1)^2}\sum^\inf_{k=0} Y_k[n] $

As the system is linear, thus the input would be

$ \frac{1}{(k+1)^2}\sum^\inf_{k=0} X_k[n] $

$ =\frac{1}{(k+1)^2}\sum^\inf_{k=0} \delta[n-k] $