Prove with induction that $a_n = 2^n-1$

Question

I have $a_2=3, a_3=7, a_4=15$ and $a_n = 1+2a_{(n-1)}$

how do I use proof by induction to prove that $a_n = 2^n -1 ?$

I'm aware of the steps with showing that its true for $n=1$ and assuming true for $n=k$ and showing for $n=k+1$ but I just don't understand how to actually do the two last steps..

Answer 1

We use induction. Now, we already know that $a_1=\frac{a_2-1}{2}=\frac{3-1}{2}=1=2^1-1$.

Assume that we already know the statement is true for $n=k$, i.e. we know $a_k=2^k-1$.

Then, since $a_{k+1}=2a_k+1=2(2^k-1)+1=2^{k+1}-1$, we get that the statement is true for $n=k+1$ as well. This completes the induction.

Answer 2

$$a_n+1=2\left(a_{n-1}+1\right)=2^2\left(a_{n-2}+1\right)=...=2^{n-2}\left(a_{2}+1\right)=2^n$$ $$\therefore a_n=2^n-1$$