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3 years ago

Find the sum of the geometric series 1+ 0.8 + 0.8^2 + 0.8^3 + ... + 0.8^19

Mathematics

2 answers:

lubasha [3.4K]3 years ago

6 0

<h3>Answer: Approximately 4.94235392476964</h3>

Work Shown:

a = first term = 1

r = 0.8 = common ratio

We know the common ratio is 0.8 because we multiply this common ratio by each term to get the next term. The easiest to spot this is going from term1 to term2. Put another way, term2/term1 = 0.8/1 = 0.8 is our common ratio.

The general form of the nth term of a geometric sequence is $a(r)^{n-1}$ and because the last term has an exponent of 19, this means n-1 = 19 solves to n = 20. Therefore, we are summing exactly n = 20 terms here.

Use the formula below to take a shortcut rather than painstakingly find all 20 terms and then summing them.

$S = \frac{a*(1-r^n)}{1-r}\\\\S = \frac{1*(1-0.8^20)}{1-0.8}\\\\S \approx \frac{1*(1-0.01152921504607)}{1-0.8}\\\\S \approx \frac{1*(0.98847078495393)}{0.2}\\\\S \approx \frac{0.98847078495393}{0.2}\\\\S \approx 4.94235392476964\\\\$

This value is approximate.

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To check your answer, you could type the following into your calculator

1+0.8+0.8^2+0.8^3+0.8^4+0.8^5+0.8^6+0.8^7+0.8^8+0.8^9+0.8^10+0.8^11+0.8^12+0.8^13+0.8^14+0.8^15+0.8^16+0.8^17+0.8^18+0.8^19

Though as you can see, this is very tedious to type in. This is why the formula in the previous section above is preferred.

Oxana [17]3 years ago

3 0

Answer:idk I still trying to find it

Step-by-step explanation:

Sorry

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Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

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$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

If solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.9=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by: