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Let $a_{0}=1$ and $a_{1}=7$ and let for all $n\geq 1$, $a_{n+1}=2a_{n}+3a_{n-1}$. Show that for all $n\geq 0$, $a_{n}=2.3^{n}-\left( -1\right) ^{n}$.

Proof. let $P_{n}=a_{n}=2.3^{n}-\left( -1\right) ^{n}$.

Initial Step. $P_{0}=1$ and $P_{1}=7$. True.

Inductive Step. $P_{n+1}=a_{n+1}=2a_{n}+3a_{n-1}=2\left( 2.3^{n}-\left( -1\right) ^{n}\right) +3\left( 2.3^{n-1}-\left( -1\right) ^{n-1}\right)=2.3.3^{n}+\left( -1\right) ^{n}=2\cdot 3^{n+1}-\left( -1\right) ^{n+1} $

Therefore, $P_{n+1}$ holds.

Can you check my proof?

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    must the proof use induction? – 2017-01-21
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    Yes. This question is in the section of induction exercises – 2017-01-21
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    @Dr.SonnhardGraubner Did you check my proof? So, Do you know different proof? – 2017-01-21
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    the proof seems correct for me – 2017-01-21
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    yes i know a different proof of this formula – 2017-01-21
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    @Dr.SonnhardGraubner Can you write here? – 2017-01-21
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    When you write $2.3^n$, do you really mean $2\cdot 3^n$? Raising $2.3=\frac{23}{10}$ to successive powers will not produce an integer sequence! – 2017-01-21
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    @HenningMakholm Yes. You are right. How should I use notation as latex? – 2017-01-21
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    No this proof is not correct, actually one cannot use this recursion hypothesis $P_n$ since it involves only $a_n$ while $a_{n+1}$ involves $a_n$ **AND** $a_{n-1}$. Can you define and use another recursion hypothesis $P_n$? (Friendly advice: If you want to know why asking user Graubner to check the validity of a proof, any proof, is a bad idea, check some of their answers. That is, some answer not yet deleted for being utter nonsense...) – 2017-02-06
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    @Did Could you explain of which other recursion hypothesis $P_n$ you are thinking? Because I don't see how induction can help with this problem. Can't I just set $a_n := 2 \cdot 3^n$ and then the calculation of the OP shows that $a_{n+1} = 2 a_{n} + 3 a_{n-1}$? This proves that $2 \cdot 3^n$ solves $a_{n+1} = 2 a_{n} + 3 a_{n-1}$ for all $n$. I think the term "induction" is not appropriate but the calculation shows that $2 \cdot 3^n$ is a solution. – 2017-02-07
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    Except that $a_n=2\cdot3^n$ does not fit the conditions $a_0=1$ and $a_1=7$ so $a_n=2\cdot3^n$ is **not** a solution. To prove recursively that $a_n=2\cdot3^n-(-1)^n$, consider the assertion: $$P_n:\ a_{n+1}=2\cdot3^{n+1}-(-1)^{n+1}\ \&\ a_n=2\cdot3^n-(-1)^n$$ then proceed as usual: check that $P_0$ holds and show that, for every $n\geqslant0$, $P_n$ implies $P_{n+1}$. – 2017-02-07
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    @Did Sorry, this was a typo, of course I forgot to subtract $(-1)^n$. But my question remains the same. Define the function $f(n) := 2 \cdot 3^n - (-1)^n$. Then $f(n+1) = 2 f(n) + 3f(n-1)$ for all $n$. Why do I need induction? – 2017-02-07
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    To show that $f(n+1)=2f(n)+3f(n-1)$ is to perform the part [$P_n$ implies $P_{n+1}$] of the proof by induction (that is, with the correct induction hypothesis $P_n$ explained in my comment). Once this is done, one still needs to check that $f(0)$ and $f(1)$ are as they should -- a task that should evoke the initialization step of the recursion proof... At the end of the day, I suggest to go for the formalized version of the proof explained in my comment, if only for safety's sake. Later on, with hundreds of proofs under your belt, more informal versions might be okay, but not now... – 2017-02-07

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we have $$a_{n+1}=2a_n+3a_{n-1}$$ and $$a_0=1,a_1=7$$ setting $$a_n=q^n$$ then we get the equation $$q^2-2q-3=0$$ with the solutions $$q_1=3,q_2=-1$$ thus you will get $$a_n=C_1\cdot 3^n+C_2\cdot (-1)^n$$ you can compute $$C_1,C_2$$ using the conditions $$a_0=1,a_1=7$$