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How would I solve this recurrence relation? $$a_n=4a_{n-1}+3$$

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  • Welcome to Math.SE! You'll find that you get significantly more (and better) help here if you start by showing us what you've tried and where you're getting stuck. That means people can address your confusion instead of just answering the question for you. – Nick Peterson Nov 02 '21 at 18:28
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    Write the first few $a_n$ in terms of $a_0$, conjecture a general result, then prove it by induction. – J.G. Nov 02 '21 at 18:28

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Put $u_n=a_n+1$ then $$ u_{n+1} = a_{n+1}+1=4a_n+3+1=4\left(a_n+1\right)=4u_n $$ Then $u_n=\left(a_0+1\right)4^n$ and $$ a_n=\left(a_0+1\right)4^n-1 $$

Just a tip : If u have $a_{n+1} = \alpha a_n + \beta$, you can solve it by considering the auxiliary sequence $u_n = a_n - \frac{\beta}{1-\alpha}$ if $\alpha \ne 1$.

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