Quessyrutty6w

2022-07-15

Find a explicit formula for the recursion
$\begin{array}{rl}{a}_{1}& =3\\ {a}_{n}& =5{a}_{n-1}+12\end{array}$
So i found
$\begin{array}{rl}{a}_{1}& =3\\ {a}_{2}& =27\\ {a}_{3}& =147\\ {a}_{4}& =747\end{array}$

Monica Dennis

Expert

Step 1
Try for characteristic equation:
${a}_{n}=5{a}_{n-1}+12\phantom{\rule{0ex}{0ex}}r=5\phantom{\rule{0ex}{0ex}}{a}_{n}={c}_{1}{r}^{n}+{c}_{2}\phantom{\rule{0ex}{0ex}}{a}_{n}={c}_{1}\left(5{\right)}^{n}+{c}_{2}\phantom{\rule{0ex}{0ex}}$
Step 2
apply now ${a}_{1}=3,{a}_{2}=27$ to find ${c}_{1},{c}_{2}$
${a}_{1}=3={c}_{1}5+{c}_{2}\phantom{\rule{0ex}{0ex}}{a}_{2}=27={c}_{1}25+{c}_{2}\phantom{\rule{0ex}{0ex}}{a}_{n}=\frac{6}{5}\left(5{\right)}^{n}-3\phantom{\rule{0ex}{0ex}}{a}_{n}=6\left(5{\right)}^{n-1}-3$

owsicag7

Expert

Step 1
If substitute ${a}_{n-2}$, ${a}_{n-3}$, and so on, we have:
${a}_{n}={5}^{k}\cdot {a}_{n-k}+12\cdot \sum _{i=0}^{k-1}{5}^{i}$
${a}_{n}={5}^{k}\cdot {a}_{n-k}+12\cdot \frac{{5}^{k}-1}{5-1}$
${a}_{n}={5}^{k}\cdot {a}_{n-k}+3\cdot \left({5}^{k}-1\right)$
Step 2
Let: $k=n-1$, then $n-k=1$
${a}_{n}={5}^{n-1}\cdot {a}_{1}+3\cdot \left({5}^{n-1}-1\right)$

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