If $\$ 4000$ is invested at $5 \%$ interest, compounded annually, then after $n$ years the investment is worth $a_{n}=4000(1.05)^{n}$ dollars.
(a) Find the first five terms of the sequence $\left\{a_{n}\right\}$. (Round your answers to the nearest cent.)
$$
\begin{array}{l}
a_{1}=\$ 4200 \
a_{2}=\$ 4400 \
a_{3}=\$ 4600 \
a_{4}=\$ 4900 \
a_{5}=\$ 5
\end{array}
$$
(b) Is the sequence convergent or divergent?
convergent
divergent

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the investment formula $a_n = 4000(1.05)^n$, we need to find the first five terms of the sequence. ‖ ‖ step 2 ⋮ Calculate $a_1$: $a_1 = 4000(1.05)^1 = 4000 imes 1.05 = 4200$. ‖ ‖ step 3 ⋮ Calculate $a_2$: $a_2 = 4000(1.05)^2 = 4000 imes 1.1025 = 4410$. ‖ ‖ step 4 ⋮ Calculate $a_3$: $a_3 = 4000(1.05)^3 = 4000 imes 1.157625 = 4630.50$. ‖ ‖ step 5 ⋮ Calculate $a_4$: $a_4 = 4000(1.05)^4 = 4000 imes 1.21550625 = 4862.03$. ‖ ‖ step 6 ⋮ Calculate $a_5$: $a_5 = 4000(1.05)^5 = 4000 imes 1.2762815625 = 5105.13$. ‖ ‖ step 7 ⋮ To determine if the sequence is convergent or divergent, we analyze the limit as $n$ approaches infinity. ‖ ‖ step 8 ⋮ Since $1.05 > 1$, the term $(1.05)^n$ grows without bound as $n o \infty$. ‖ ‖ step 9 ⋮ Therefore, $\lim_{n o \infty} 4000(1.05)^n = \infty$, indicating that the sequence is divergent. ‖ ∻Answer∻ ⚹ The first five terms of the sequence are $4200, 4410, 4630.50, 4862.03, 5105.13$. The sequence is divergent. ⚹ ∻Key Concept∻ ⚹Exponential Growth⚹ ∻Explanation∻ ⚹The sequence $a_n = 4000(1.05)^n$ grows exponentially because the base $1.05$ is greater than $1$, leading to divergence as $n$ approaches infinity.⚹◊What is the formula for the nth term of the sequence given by $a_{n}=4000(1.05)^{n}$?⍭ Generate me a similar question◊

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