Solution:

Consider the accumulator, We define $x_1[n]=x\left[n-n_0\right]$. To show time invariance, we solve for both $y\left[n-n_0\right]$ and $y_1[n]$ and compare them to see whether they are equal. First,

Next, we find,

$$ y\left[n-n_0\right]=\sum_{k=-\infty}^{n-n_0} x[k] $$

$$ \begin{aligned} y_1[n] & =\sum_{k=-\infty}^n x_1[k] \\ & =\sum_{k=-\infty}^n x\left[k-n_0\right] . \end{aligned} $$

Substituting the change of variables $k_1=k-n_0$ into the summation gives,

$$ y_1[n]=\sum_{k_1=-\infty}^{n-n_0} x\left[k_1\right]=y\left[n-n_0\right] . $$

Thus, the accumulator is a time-invariant system. The following example illustrates a system that is not time-invariant.