Given recursively defined sequence:
$$a_1=1, a_n=2a_{n-1}+3, n\geq 2$$
prove $a_n=2^{n+1}-3$.
Let's prove this by induction:
Base case: for $n=1, a_1=2^2-3=1$ so the statements holds.
Let's assume that for some $n\in\Bbb N, a_n=2^{n+1}-3$.
Then $$a_{n+1}=2a_n+3=2(2^{n+1}-3)+3=2^{n+2}-3$$
Q.E.D.
I'm not sure whether I'm allowed to put $a_{n+1}=2a_n+3$?
Maybe I should've assumed $a_{n-1}=2^n-3$ and then get $$a_n=2a_{n-1}+3=2(2^n-3)+3=2^{n+1}-3.$$
Which one of these is correct?
Induction is much easier for $b_n=a_n+3$.
Indeed, $b_{n}=a_n+3=2a_{n-1}+3+3=2(b_{n-1}-3)+6=2b_{n-1}$, from which it follows that $b_n=b_1 2^{n-1}=4 \cdot 2^{n-1} = 2^{n+1}$.