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What would be the best way in finding a general term $a_{n}$, $n \geq 2$ for the recursive sequence $a_{n} = 3a_{n - 1} + 1$, where $a_{1} = 1$.

user2850514
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2 Answers2

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Let $$b_n=a_n+\frac 1 2$$ then

$$b_{n}=3a_{n-1}+\frac 3 2=3\left(a_{n-1}+\frac 1 2\right)=3b_{n-1}$$ so $(b_n)$ is a geometric sequence. Express $b_n$ using $n$ and the first term $b_1$ and then you find $a_n$.

Edit We have $$b_n=3^{n-1}b_1=3^{n-1}\left(a_1+\frac 1 2\right)=\frac{3^n}{2}$$ so $$a_n=b_n-\frac 1 2=\frac{3^n-1}{2}$$

  • Can you explain why you let $b_{n} = a_{n} + \frac{1}{2}$? – user2850514 Dec 29 '13 at 20:09
  • $(a_n)$ is called arithmetico-geometric sequence. The general method is to find $\ell$ such that $\ell=3\ell+1$ i.e. replace $a_n$ and $a_{n-1}$ by $\ell$ and then $b_n=a_n-\ell$ is a geometric sequence. –  Dec 29 '13 at 20:13
  • Ok, so now we have this, how do we find a general term for $a_{n}$? – user2850514 Dec 29 '13 at 20:26
  • You find the expression of $b_n$ (since it's a geometric sequence) and then $a_n=b_n-\frac 1 2$. –  Dec 29 '13 at 20:27
  • @user2850514 I edited my answer. –  Dec 29 '13 at 20:34
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Well, note that $a_n = 3 a_{n-1} + 1 = 9 a_{n-2} + 4 = 27 a_{n-3} + 13,$ etc. So, you conjecture that $a_n = 3^{n-1} a_1 + \sum_{i=0}^{n-2} 3^i=3^{n-1} + \sum_{i=0}^{n-2} 3^i.$ Now, you check that this works by induction.

Asinomás
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Igor Rivin
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