# Finite Geometric Series: Derivation and Problems (Grades 11-12)

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## Finite Geometric Series: Derivation and Problems

1.

The following algebraic work shows the derivation of the formula for the sum of a finite geometric series, [math]S_n = a((1-r^n)/(1-r))[/math]. Answer the following questions to give the reasons or justification for each step in the work.

[math](1) \ \ S_n = a + ar + ar^2 + ... + ar^(n-2) + ar^(n-1) [/math]

[math](2) \ \ S_n r = ar + ar^2 + ar^3 + ... + ar^(n-1) + ar^n [/math]

[math](3) \ \ S_n - S_n r = a - ar^n[/math]

[math](4) \ \ S_n (1-r) = a(1-r^n) [/math]

[math](5) \ \ S_n = a ((1-r^n)/(1-r))[/math]

[math](1) \ \ S_n = a + ar + ar^2 + ... + ar^(n-2) + ar^(n-1) [/math]

[math](2) \ \ S_n r = ar + ar^2 + ar^3 + ... + ar^(n-1) + ar^n [/math]

[math](3) \ \ S_n - S_n r = a - ar^n[/math]

[math](4) \ \ S_n (1-r) = a(1-r^n) [/math]

[math](5) \ \ S_n = a ((1-r^n)/(1-r))[/math]

A.

Where does the equation in step one come from?

- From the definition of a geometric series.
- From a well known polynomial identity.
- It is an assumption.
- It is simply a new and arbitrary definition of [math]S_n[/math].

B.

How does one arrive at step 2?

- It is a new definition for [math]S_n[/math].
- Adding a factor of [math]r[/math] to each term.
- By multiplying each side of the equation in step one by [math]r[/math].
- Adding a similar, but larger, finite geometric series.

C.

The equation in the third step is found by subtracting the equation from step 2 from the equation in step 1. What happens to all the terms on the right hand side?

- They are ignored, because they are all going to be much smaller than either [math]a[/math] or [math]ar^n[/math], depending on whether [math]r[/math] is greater or less than 1.
- All but two of them are eliminated by subtraction. Aside from the first term of the first equation and the last term of the second equation, each term in the first equation has an equal term in the second equation, and thus they become n - 2 zeros.
- Using the factor theorem, they all cancel out except for [math]a[/math] and [math]ar^n[/math].
- Dividing both sides of the equation by [math]ar, ar^2, ... , ar^(n-1)[/math], they all cancel out.

D.

How does one go from step 3 to step 4?

- Use polynomial long division.
- Apply the fundamental theorem of algebra.
- Factor out common factors on both sides.
- Multiply each side by [math](1-r)[/math].

E.

Which of the following best explains how one arrives at the equation in step 5?

- By stating the definition of a finite geometric sequence.
- Subtract [math](1-r)[/math] from both sides of the equation in step 4, and then factor out common factors on the left hand side of the resulting equation.
- Factor [math](1-r)[/math] from both sides of the equation in step 4.
- Divide both sides of the equation in step 4 by [math](1-r)[/math].

F.

Are there any restrictions one should have added at the beginning of this derivation? Choose all that apply.

- No, this correct as is.
- Yes, this derivation is only valid for [math]a>0[/math].
- Yes, the formula found is only true if [math]|r|<1[/math].
- Yes, it must be stated that [math]r!=1[/math].

2.

Using the polynomial property for [math]x^n-1[/math], namely that [math](x^n - 1)/(x-1) = x^(n-1) + x^(n-2) + ... + x^2 + x + 1[/math], derive the formula for a finite geometric series.

3.

A ball rolling down an incline travels 2 meters in the first second, 4 meters the second, and 8 meters the third second. What is the total distance it will it roll after 10 seconds?

- 256 meters
- 2,046 meters
- 20 meters
- 1,024 meters

4.

A business pays $300 to rent a piece of office equipment for one year. The rent was increased by 15% each year thereafter. How much will the company pay for the first four years it rents the equipment, to the nearest dollar?

- $456
- $1,498
- $1,153
- $853

5.

Aaron is giving out Halloween candy. He notices that when he starts giving out candy there are a lot of children coming to his house, but this number decreases as the night goes on. If he gives out about 100 pieces of candy in the first 15 minutes, but every subsequent 15 minutes he hands out about 85% less candy, how much candy does he give out in total, if he starts at 6:00 p.m. and stops at 8:30 p.m.?

- 535 pieces
- 512 pieces
- 555 pieces
- 55,128 pieces

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