Ttdfjt

$y[n]=Tx[n]=displaystylesum_k=n-n_0^n+n_0 x[k]$

it is some sort of moving summer which computes $n^textth$ output sample by adding all samples lying within length $n_0$ around some point $n$ on $k$ -axis (where k is dummy variable) so even if we shift input by amount $n_0$ the length of interval of summation remain unchanged which is same as if we're shifting output by $n_0$ from which it follows system is Time invariant but how do we prove it mathematically .

my work so far

shifting input by $n_0$ , i.e, $Tx[n-n_0]=displaystylesum_k=n-n_0^n+n_0 x[k-n_0]$

now, $kto k+n_0$ then,

LHS$=displaystylesum_k=n-2n_0^n x[k]neq y[n-n_0]$

here i'm getting the answer Time variant. can anyone help ?

You are overloading the symbol $n$ which doesn't mean the same everywhere in your calculations. Instead, write down what the sequence of input values is and the corresponding sequence of output values when the input is $x$. DO NOT use the symbol $n$ anywhere ($n_0$ is OK to use). So, your answer should look like

beginalign
y[-2] &= cdots\
y[-1] &= cdots\
y[0] &= x[-n_0] + x[-n_0+1] + cdots + x[0] + x[1] + cdots x[n_0]\
y[1] &= x[-n_0+1] + cdots + x[0] + x[1] + cdots x[n_0+1]
endalign
etc.

Now, shift $x$ by $k$ places (say $k=1$ for starters) and call the result $hat x$. Repeat the above computation for the input $hat x$ and call the result $hat y$. Is $hat y$ the same as $y$ except shifted by $k$ places? Repeat for $k+1$ etc until you have managed to convince yourself that regardless of what $k$ you choose, shifting $x$ by $k$ places results in $y$ also shifting by $k$ places.

In addition to Dilip's hands on answer, let me correct the theoretical path that you wanted to follow but failed.

The input output relation of the system is given by :

$$ y[n] = T x[n] = sum_k=n-n_0^n+n_0 x[k] tag1$$

First, shift the input $x[n]$, by integer $d$, denoted as
$$x_d[n] = x[n-d] tag2$$
and see its effect on the output $y[n]$; denoted as $y_d[n]$ :

$$ y_d[n] = Tx_d[n] = sum_k=n-n_0^n+n_0 x_d[k] tag3 $$

Replace $x_d[n]$ with $x[n-d]$ and $x_d[k]$ with $x[k-d]$ according to Eq.2.
$$ y_d[n] = Tx[n-d] = sum_k=n-n_0^n+n_0 x[k-d] tag4 $$

Second, shift the output $y[n]$ by $d$ and see its effect on the sum:

$$y[n-d] = sum_k=n-d-n_0^n-d+n_0 x[k] tag5 $$

Now, make a manipulation of dummy summation index $k$ in either of Eqs. 4 or 5 to see that they are the same; i.e.,

For example, let $m = k+d$ and place it into Eq.5:
$$y[n-d] = sum_ m = n-n_0^n+n_0 x[m-d] tag6 $$

and simply reset the dummy variable of summation from $m$ back to $k$ in Eq.6, to convince yourself that Eq.6 becomes Eq.4:

$$y[n-d] = sum_ k = n-n_0^n+n_0 x[k-d] = y_d[n] tag7 $$

Finally, Eq.7 declares that $y[n-d] = y_d[n]$; the system is Time-Invariant.