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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?

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    It looks correct ... it's a classical induction at work, either you assume step $n$ and prove step $n+1$ or assume step $n-1$ and prove step $n$, not a big difference. – 2017-01-08
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    Observe that $n$ just represent a natural number, no matter what natural number is. You can use, instead of $n$, $n+1$or $n-2$, etc... if these quantities represent an undefined natural number. After you check that the induction hypothesis hold for the next natural number, named [the successor](https://en.wikipedia.org/wiki/Successor_function). The proof by induction is based in the [axiom schema of induction](https://en.wikipedia.org/wiki/Mathematical_induction#Axiom_of_induction), that is an essential part to define the natural numbers in FOL. – 2017-01-08

1 Answers 1

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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 $b_n=b_1 2^{n-1}=4 \cdot 2^{n-1} = 2^{n+1}$.