Question

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

Answer 1

Answer: Approximately 4.94235392476964

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.

Answer 2

Answer:idk I still trying to find it

Step-by-step explanation:

Sorry