280. What is the sum of the series $\sum_{n=1}^{\infty} \frac{(2^{2n+1})}{5^{n-1}}$?

A. 40
B. 20
C. 10
D. 8

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281. The population of a certain country is 80 million with a growth rate of 2% per year.

Given $(0.02)^9 = 5.12 \times 10^{-16}$, $(1.02)^9 = 1.19$
$(0.02)^{10} = 1.024 \times 10^{-17}$, $(1.02)^{10} = 1.22$

Which of the following is the best approximation of the population (in million) after 10 years?

A. 81.9
B. 86.8
C. 95.2
D. 97.6

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282. Which one of the following represents a geometric sequence?

A. $3, 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$
B. $\frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, -\frac{1}{5}, \frac{1}{6}, \dots$
C. $1, 3, 6, 10, 15, \dots$
D. $-3, 6, -9, 12, -15, \dots$

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283. What is the value of the sum $\sum_{n=1}^{\infty} \frac{(2^n + 5^n)}{10^n}$?

A. 0.325
B. 1
C. $\frac{5}{4}$
D. $\frac{37}{9}$

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284. What is the sum of the series $\sum_{n=1}^{\infty} (-1)^n (3)^{-2n}$?

A. $-\frac{1}{8}$
B. -0.13
C. -0.1
D. $\frac{1}{8}$

Sum of the Series [tex]\sum_{n=1}^{\infty} \frac{(2^{2n+1})}{5^{n-1}}[/tex]:
We can start by simplifying the general term:
[tex]\frac{2^{2n+1}}{5^{n-1}} = 2 \cdot \frac{4^n}{5^n} \cdot 5 = 2 \cdot \left(\frac{4}{5}\right)^n \cdot 5
= 10 \cdot \left(\frac{4}{5}\right)^n.[/tex]
This is indeed a geometric series with first term [tex]a = 10[/tex] and common ratio [tex]r = \frac{4}{5}[/tex].
The sum [tex]S[/tex] of an infinite geometric series is given by:
[tex]S = \frac{a}{1 - r} = \frac{10}{1 - \frac{4}{5}} = 10 \times 5 = 50.[/tex]
Therefore, the sum of the series is 50.
This solution does not match any of the provided options, which suggests there may be a mistake in the options listed.

Population Growth Approximation:
The population after 10 years can be approximated using the formula for exponential growth:
[tex]P = P_0 \times (1 + r)^n[/tex]
Here, [tex]P_0 = 80[/tex] million, [tex]r = 0.02[/tex] (or 2%), and [tex]n = 10[/tex].
So we calculate:
[tex]P = 80 \times 1.22 = 97.6 \text{ million}.[/tex]
Thus, the best approximation for the population after 10 years is [tex]\text{D. } 97.6[/tex] million.

Geometric Sequence Identification:
A geometric sequence is one in which each term is a constant multiple of the previous term.
Let's evaluate the given sequences:
A. [tex]3, 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots[/tex]
The ratio [tex]r = \frac{1}{3}[/tex], as each term is [tex]\frac{1}{3}[/tex] of the previous one.
Options B, C, and D do not represent geometric sequences as they lack a constant ratio between consecutive terms.
Therefore, Option A is the correct choice.

Sum of the Series [tex]\sum_{n=1}^{\infty} \frac{(2^n + 5^n)}{10^n}[/tex]:
We can split the series into two separate geometric series:
[tex]\sum_{n=1}^{\infty} \left(\frac{2^n}{10^n} + \frac{5^n}{10^n}\right) = \sum_{n=1}^{\infty} \left(\frac{2}{10}\right)^n + \sum_{n=1}^{\infty} \left(\frac{5}{10}\right)^n.[/tex]
The sums of the separate series are:
[tex]\frac{\frac{2}{10}}{1 - \frac{2}{10}} = \frac{0.2}{0.8} = 0.25[/tex]
[tex]\frac{\frac{5}{10}}{1 - \frac{5}{10}} = \frac{0.5}{0.5} = 1[/tex]
Adding these, the total sum is [tex]0.25 + 1 = 1.25 = \frac{5}{4}[/tex].
Therefore, the correct answer for the sum of the series is Option C. [tex]\frac{5}{4}[/tex].

Sum of the Series [tex]\sum_{n=1}^{\infty} (-1)^n (3)^{-2n}[/tex]:
Let's consider the term: [tex](-1)^n \left(\frac{1}{9}\right)^n = \left(-\frac{1}{9}\right)^n[/tex].
Using the formula for an infinite geometric series, the sum is
[tex]S = \frac{a}{1 - r}, \text{ where } a = -\frac{1}{9} \text{ and } r = -\frac{1}{9}.[/tex]
Thus:
[tex]S = \frac{-\frac{1}{9}}{1 + \frac{1}{9}} = \frac{-\frac{1}{9}}{\frac{10}{9}} = -\frac{1}{10}.[/tex]
Therefore, the sum of the series is Option C. -0.1.