babeeb0oL

2021-07-30

Solve the following recurrence relation:
a)
b)
c) and ${F}_{1}=4$

Nathaniel Kramer

Expert

Step 1
According to the given information, it is required to solve the recurrence relation.
a)
${a}_{0+1}={a}_{1}=d{a}_{0}+c⇒{a}_{1}=c$
${a}_{1+1}={a}_{2}=d{a}_{1}+c⇒{a}_{2}=dc+c=\left(d+1\right)c$
${a}_{2+1}={a}_{3}=d{a}_{2}+c⇒{a}_{3}=d\left(dc+c\right)+c⇒{a}_{3}={d}^{2}c+dc+c=\left({d}^{2}+d+1\right)c$
${a}_{3+1}={a}_{4}=d{a}_{3}+c⇒{a}_{4}=d\left({d}^{2}c+dc+c\right)+c⇒{a}_{4}=\left({d}^{3}+{d}^{2}+d+1\right)c$
$\begin{array}{|c|}\hline {a}_{n}=\left({d}^{n}+{d}^{n-1}+{d}^{n-2}+\cdots +d+1\right)c\\ \hline\end{array}$
Step 2
b)
${a}_{0+1}^{3}={a}_{1}^{3}=2{a}_{0}^{3}⇒{a}_{1}^{3}=2{\left(5\right)}^{3}={2}^{1}{\left(5\right)}^{3}$
${a}_{1+1}^{3}={a}_{2}^{3}=2{a}_{1}^{3}⇒{a}_{2}^{3}=2×2{\left(5\right)}^{3}={2}^{2}{\left(5\right)}^{3}$
${a}_{2+1}^{3}={a}_{3}^{3}=2{a}_{2}^{3}⇒{a}_{3}^{3}=2×2×2{\left(5\right)}^{3}={2}^{3}{\left(5\right)}^{3}$

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