How to solve this question by using convolution?
Assuming you mean that $y[n] = h[n] * x[n]$, then 
\begin{equation}
\begin{split}
 y[n]  &amp;\stackrel{(a)}{=} h[n] * x[n]\\
 &amp;\stackrel{(b)}{=} \sum_{m=-\infty}^{\infty}
 h[m]x[n-m]\\
 &amp;\stackrel{(c)}{=} \sum_{m=-\infty}^{\infty}
 u[m]2^{n-m}u[n-m]\\
 &amp;\stackrel{(d)}{=}2^n
 \sum_{m=-\infty}^{\infty}
 u[m]2^{-m}u[n-m]\\
 &amp;\stackrel{(e)}{=}2^n
 \sum_{m=0}^{\infty}
 2^{-m}u[n-m]\\
 &amp;\stackrel{(f)}{=}2^n
 \sum_{m=0}^{n}
 2^{-m}\\
 &amp;\stackrel{(g)}{=}2^n
 \frac{1 - (\frac{1}{2})^{n+1}}{1 - \frac{1}{2}}\\
 &amp;\stackrel{(h)}{=}2^{n+1} - 1
\end{split}
\end{equation}

(a) is the convolution symbol.

(b) is using the definition of discrete convolution.

(c) replacing the quantities.

(d) $2^n$ is independent of the summation index.

(e) $u[m]=1$ if $m\geq 0$ and $0$ otherwise.

(f) if $n-m&lt;0$, or if $n&lt;m$ then the summation is zero. So we have (for non-zero quantities), $m = 0 \ldots n$.

(g) Using geometric series

(h) rearrangements.