Given:
$U(0)=4$
$U(n+1)=3+2\cdot U(n)$ ...recurrence case
$U(n)=7\cdot 2^n-3$
My Solution:
$U(n+1)=3+2\cdot U(n)$
$\frac{U(n+1)-3}{2}=U(n)$
$\frac{U(n+1)-3}{2}=7\cdot 2^n-3$
Since I know $U(0) =4$, and with that I can solve $U(1)=11$
$n = 0$
$\frac{U(0+1)-3}{2}=7\cdot2^0-3$
$4=4$ .... is this a proof?
Not really a proof, but it contains the idea.
What you're doing should be formalized with induction.
Given: $U(0)=4$, $U(n+1)=3+2U(n)$
Conjecture: $U(n)=7\cdot 2^n-3$
Base of induction: $7\cdot2^0-3=4=U(0)$. Verified true.
Induction step: Assume $U(n)=7\cdot2^n-3$. Then $$ U(n+1)=3+2U(n)=3+2(7\cdot2^n-3)=2\cdot7\cdot2^{n}+3-6=7\cdot2^{n+1}-3. $$ End.