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prisoha [69]
1 year ago
11

Investments increase exponentially by

Mathematics
2 answers:
aleksklad [387]1 year ago
8 0

Answer:

9060$ is the final amount because 7060 is the interest

STatiana [176]1 year ago
6 0

The amount of money after 45 years will be $64,060.

<h3>What is compound interest?</h3>

Compound interest is the interest on a loan or deposit calculated based on the initial principal and the accumulated interest from the previous period.

We know that the compound interest is given as

A = P(1 + r)ⁿ

Where A is the amount, P is the initial amount, r is the rate of interest, and n is the number of years.

Investments increase exponentially by about 26% every 3 years.

If you made a $2,000 investment.

Then the equation will be

\rm A = 2000 \times \left (1.26 \right )^{\frac{t}{3}}

Where t is the number of years.

Then the amount of money after 45 years will be

\rm A = 2000 \times \left (1.26 \right )^{\frac{45}{3}}

Simplify the equation, then we have

A = 2000 × (1.26)¹⁵

A = 2000 × 32.03

A = $64,060

More about the compound interest link is given below.

brainly.com/question/25857212

#SPJ1

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I think that is the right answer I hope ut helps

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Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
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Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

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