Quessyrutty6w

Answered

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

$\begin{array}{rl}{a}_{1}& =3\\ {a}_{2}& =27\\ {a}_{3}& =147\\ {a}_{4}& =747\end{array}$

$\begin{array}{rl}{a}_{1}& =3\\ {a}_{n}& =5{a}_{n-1}+12\end{array}$

So i found

$\begin{array}{rl}{a}_{1}& =3\\ {a}_{2}& =27\\ {a}_{3}& =147\\ {a}_{4}& =747\end{array}$

Answer & Explanation

Monica Dennis

Expert

2022-07-16Added 13 answers

Step 1

Try for characteristic equation:

${a}_{n}=5{a}_{n-1}+12\phantom{\rule{0ex}{0ex}}r=5\phantom{\rule{0ex}{0ex}}{a}_{n}={c}_{1}{r}^{n}+{c}_{2}\phantom{\rule{0ex}{0ex}}{a}_{n}={c}_{1}(5{)}^{n}+{c}_{2}\phantom{\rule{0ex}{0ex}}$

Step 2

apply now ${a}_{1}=3,{a}_{2}=27$ to find ${c}_{1},{c}_{2}$

${a}_{1}=3={c}_{1}5+{c}_{2}\phantom{\rule{0ex}{0ex}}{a}_{2}=27={c}_{1}25+{c}_{2}\phantom{\rule{0ex}{0ex}}{a}_{n}=\frac{6}{5}(5{)}^{n}-3\phantom{\rule{0ex}{0ex}}{a}_{n}=6(5{)}^{n-1}-3$

Try for characteristic equation:

${a}_{n}=5{a}_{n-1}+12\phantom{\rule{0ex}{0ex}}r=5\phantom{\rule{0ex}{0ex}}{a}_{n}={c}_{1}{r}^{n}+{c}_{2}\phantom{\rule{0ex}{0ex}}{a}_{n}={c}_{1}(5{)}^{n}+{c}_{2}\phantom{\rule{0ex}{0ex}}$

Step 2

apply now ${a}_{1}=3,{a}_{2}=27$ to find ${c}_{1},{c}_{2}$

${a}_{1}=3={c}_{1}5+{c}_{2}\phantom{\rule{0ex}{0ex}}{a}_{2}=27={c}_{1}25+{c}_{2}\phantom{\rule{0ex}{0ex}}{a}_{n}=\frac{6}{5}(5{)}^{n}-3\phantom{\rule{0ex}{0ex}}{a}_{n}=6(5{)}^{n-1}-3$

owsicag7

Expert

2022-07-17Added 2 answers

Step 1

If substitute ${a}_{n-2}$, ${a}_{n-3}$, and so on, we have:

${a}_{n}={5}^{k}\cdot {a}_{n-k}+12\cdot \sum _{i=0}^{k-1}{5}^{i}$

${a}_{n}={5}^{k}\cdot {a}_{n-k}+12\cdot \frac{{5}^{k}-1}{5-1}$

${a}_{n}={5}^{k}\cdot {a}_{n-k}+3\cdot ({5}^{k}-1)$

Step 2

Let: $k=n-1$, then $n-k=1$

${a}_{n}={5}^{n-1}\cdot {a}_{1}+3\cdot ({5}^{n-1}-1)$

If substitute ${a}_{n-2}$, ${a}_{n-3}$, and so on, we have:

${a}_{n}={5}^{k}\cdot {a}_{n-k}+12\cdot \sum _{i=0}^{k-1}{5}^{i}$

${a}_{n}={5}^{k}\cdot {a}_{n-k}+12\cdot \frac{{5}^{k}-1}{5-1}$

${a}_{n}={5}^{k}\cdot {a}_{n-k}+3\cdot ({5}^{k}-1)$

Step 2

Let: $k=n-1$, then $n-k=1$

${a}_{n}={5}^{n-1}\cdot {a}_{1}+3\cdot ({5}^{n-1}-1)$

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