# replace input signal with δ will have the impulse response of the function?

Is my impulse response right? By definition,the impulse response is the output when the input is a impulse signal,so

$$y[n]=\sum\limits ^{n}_{k=-\infty}\frac{1}{2^{n-k}}\ x[k]$$,the impulse response of $$y[n]=\sum\limits ^{n}_{k=-\infty}\frac{1}{2^{n-k}}\ \delta[k]$$ ,Is my thinking right?

Of course, if you have a formula describing the system's response to an arbitrary input $$x[n]$$, you will obtain the response to an impulse if you choose $$x[n]=\delta[n]$$.
You can also find that simple expression from your second formula by realizing that $$\delta[k]$$ is only non-zero for exactly one value of $$k$$. Depending on the value of $$n$$, that value may or may not be reached within the given summation limits.