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Inessa [10]
3 years ago
13

Suppose that $2700 is borrowed for four years at an interest rate of 3% per year, compounded continuously. Find the amount owed,

assuming no payments are made until the end.
Do not round any intermediate computations, and round your answer to the nearest cent.
Mathematics
1 answer:
Elan Coil [88]3 years ago
3 0
you have to make a around the year with 2700$
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Answer:

3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n

4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n

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Step-by-step explanation:

Given

See attachment for complete question

Required

Match equivalent expressions

Solving (a):

3 + 12 + 48 + 192 + 768

The expression can be written as:

3 \to 3*4^{0 --- 0

12 \to 3 * 4^{1 ---- 1

48 \to 3 * 4^{2 --- 2

192 \to 3 * 4^{3 ---- 3

768 \to 3 * 4^{4 ---- 4

For the nth term, the expression is:

Term = 3 * 4^{n ---- n

So, the summation is:

3 + 12 + 48 + 192 + 768 = \sum\limits^4_{n=0} 3 * 4^n

Solving (b):

4 + 32 + 256 + 2048 + 16384

The expression can be written as:

4 \to 4 * 8^0 --- 0

32 \to 4 * 8^1 ---- 1

256 \to 4 * 8^2 --- 2

2048 \to 4 * 8^3 ---- 3

16384 \to 4 * 8^4 ---- 4

For the nth term, the expression is:

Term \to 4 * 8^n ---- n

So, the summation is:

4 + 32 + 256 + 2048 + 16384 = \sum\limits^4_{n=0} 4 * 8^n

Solving (c):

2 + 6 + 18 + 54 + 162

The expression can be written as:

2 \to 2 * 3^0 --- 0

6 \to 2 * 3^1 ---- 1

18 \to 2 * 3^2 --- 2

54 \to 2 * 3^3 ---- 3

162 \to 2 * 3^4 ---- 4

For the nth term, the expression is:

Term \to 2 * 3^n ---- n

So, the summation is:

2 + 6 + 18 + 54 + 162 = \sum\limits^4_{n=0} 2* 3^n

Solving (d):

3 + 15 + 75 + 375 + 1875

The expression can be written as:

3 \to 3 * 5^0 --- 0

15 \to 3 * 5^1 ---- 1

75 \to 3 * 5^2 --- 2

375 \to 3 * 5^3 ---- 3

1875 \to 3 * 5^4 ---- 4

For the nth term, the expression is:

Term \to 3 * 5^n ---- n

So, the summation is:

3 + 15 + 75 + 375 + 1875 = \sum\limits^4_{n=0} 3* 5^n

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