How did we get this final answer in example in fig attached. I understand that delta(n) is a DT unit impulse where it is 1 at n = 0 and zero elsewhere. I understand that u[n] is a unit-step where it is 1 for n>=0 and 0 elsewhere.
Draw the three discrete time signals (shown on the left side) on top of one another. You will see the summation leading to $\delta[n]$ and $\delta[n-1]$ term as follows. For $n=0$, only the first term survives and hence $3\delta[n]$. For the second term, put $n=1$ that makes $3/2+2=7/2$ at index $1$.
Then, obviously third term extends in $u[n-2]$ form as a summation from all three terms at each subsequent index. This is because $n=2$ brings $3/4$ from the first term, $1$ from the second term and $1$ from the third term, totaling to $11/4$ onwards.