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aniked [119]
4 years ago
5

PLZ PLZ PLZ Help!!! Consider the infinite series defined by

Mathematics
1 answer:
Bad White [126]4 years ago
7 0

Answer:

(a)

r =\lim_{n \to \infty} |\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}| =  \lim_{n \to \infty} \frac{n+1}{4} = \infty

(b)

if r<1, then it converges absolutely

if r>1, then it diverges

if r=1, then the test is inconclusive

Step-by-step explanation:

The ratio test is:

\lim_{n \to \infty} |\frac{x[n+1]}{x[n]}| = L

if L<1, then it converges absolutely

if L>1, then it diverges

if L=1, then the test is inconclusive

I'm assuming that they used r instead of L. Anyway, to do the ratio test, you plug in n+1 into the equation and make that numerator.

Then you plug in n into the same equation and make that your denominator.

In our case, our equation is \frac{2n!}{2^{2n}}.

So lets plug in n+1 and make that our numerator.\frac{2(n+1)!}{2^{2(n+1)}}

Now lets plug in n and make that our denominator. \frac{2n!}{2^{2n}}

Now set up the ratio test.

\lim_{n \to \infty} |\frac{x[n+1]}{x[n]}| = L

\lim_{n \to \infty} |\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}| = L

if you simplify this, you would get (view image for the steps)

\lim_{n \to \infty} |\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}| =  \lim_{n \to \infty} \frac{n+1}{4} = \infty

L = infinity, which is > 1, therefore this series diverges

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A credit card company charges 18.6% percent per year interest. Compute the effective annual rate if they compound, (a) annualy,
Darina [25.2K]

Answer:

a) Effective annual rate: 18.6%

b) Effective annual rate: 20.27%

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d) Effective annual rate: 20.44%

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In the special case the interest rate is compounded continuously, the interest is given by

I=Ce^{rt}-C

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The effective annual rate is 18.6%

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I=C(1+\frac{0.186}{12})^{12}-C=C(1.2027)-C=C(0.2027)

The effective annual rate is 20.27%

(c) Daily

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I=C(1+\frac{0.186}{365})^{365}-C=C(1.2043)-C=C(0.2043)

The effective annual rate is 20.43%

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I=Ce^{0.186}-C=C(1.2044)-C=C(0.2044)

The effective annual rate is 20.44%

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3 years ago
Please help not sure if baby has any correct ​
Elodia [21]
So I think maybe the example for the first problem created some confusion, and you may want to have your child take another look.

If we numbered the top half 1 to 9 going from left to right, numbers 1, 2, 6, 7 and 8 are right. They worked because the top number (numerator) perfectly fit into the bottom number (denominator). This is not true for the rest.

The key is to find the largest number that you can think of that will go into BOTH the top and the bottom evenly.

So for number 3: 18/24; 18 does not fit evenly into 24. The highest number that will fit into both is 6, so you divide both top and bottom by 6 and your answer is 3/4.

Number 4: 45/54; the highest number that goes into both is 9, so you divide both top and bottom by 9 and your answer is 5/6.

Number 5: 55/66; the highest number that can go into both is 11, so the answer is 5/6.

Is this making sense?

The bottom, numbering 1 to 9 from left to right. The correct ones are 1, 6 and 9.

For 2: 14/41 is about 15/40. Both can be divided by 5, so the answer is 3/8.

For 3: 20/81 is about 20/80, and 2/8 is close, but can still be divided by 2, so the answer is 1/4.

For 4: 24/49 is closer to 25/50 than 20/50. 25/50 can be divided by 25, so the answer is 1/2.

For 5: it was all correct, but the answer can be further reduced from 2/8 to 1/4.

For 7: 23/72 is about 25/75, and 25 goes into both, so it reduces to 1/3.

For 8: 13/21 is about 15/20, and 5 goes into both, so the answer is 3/4.

As your child continues to learn this, remember that if he or she gets an answer like 2/6 or 6/12, they should ask themselves if they can further reduce the fractions- 2/6 reduces to 1/3, and 6/12reduces to 1/2. I know it's confusing, but they do get the hang of it with practice
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3 years ago
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