Calculating sum of a discrete signal
One way to see it is to realize that $x[n]$ can be rewritten as the sum of 3 impulses:

$$x[n]=\delta[n]+\delta[n-1]+\delta[n-2]$$

Since $h[n]*\delta[n-k]=h[n-k]$ for arbitrary integer $k$ you get

$$h[n]*x[n]=h[n]+h[n-1]+h[n-2]=0.5^nu[n]+0.5^{n-1}u[n-1]+0.5^{n-2}u[n-2]$$

Evaluating this expression yields

$$\begin{array}{lr}0&amp;\quad n&lt;0\\
0.5^0=1&amp;\quad n=0\\
0.5^1+0.5^0=1.5&amp;\quad n=1\\
0.5^n+0.5^{n-1}+0.5^{n-2}= &amp;\\
=0.5^{n-2}(0.5^2+0.5+1)=&amp;\\
=1.75\cdot0.5^{n-2}&amp;n\ge 2
\end{array}$$

which is equivalent to the result you've given.