Yes this question can be solved by geometric progression as:
$T(2^k)
\\= 3T(2^{k-1}) + 1
\\= 3^2T(2^{k-2}) + 3 + 1$
Similarly after $k$ times
$= 3^kT(2^{k-k}) + (3^{k-1}+\dots+3+1)
\\= 3^kT(1) + \frac{3^k-1}{2}
\\= 3^k + \frac{3^k-1}{2}
\\=\frac{3.3^k-1}{2}
\\=\frac{3^{k+1}-1}{2}$