Finite geometric series

Textbook reading

The problem that opens the lesson

A ball rebounds to $70\%$ of its previous height. Starting from $10$ meters, find the total upward distance over the first $8$ rebounds.

Solution

Begin by identifying the mathematical object and the information that fixes it. Identify the first term, ratio, and number of terms, choose a stable form, and check the result against the largest term or a direct short sum. The relevant conditions are not optional bookkeeping: The case $r=1$ must be handled separately as $n a_1$ . Following that structure gives $\frac{10(0.7)(1-0.7^8)}{1-0.7}$ meters.

Why this works

The alternative form $\frac{a_1(r^n-1)}{r-1}$ is algebraically equivalent and may avoid nested negatives when $r>1$ . The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

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What this lesson is really about

A finite geometric series sums terms with a constant ratio.

Multiplying S_n by $r$ shifts every term one position. Subtracting cancels the interior terms and leaves $a_1$ and $a_1 r^n,$ giving $S_n=\frac{a_1(1-r^n)}{1-r}$ for $r\ne 1$ .

The point is not merely to reproduce a formula. A learner should be able to identify the quantities or geometric objects involved, explain why the relationship has its stated form, and recognize when the same idea appears in a graph, table, diagram, or model.

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Why the relationship works

The alternative form $\frac{a_1(r^n-1)}{r-1}$ is algebraically equivalent and may avoid nested negatives when $r>1$ .

Textbook reading

A reliable way to work

Identify the first term, ratio, and number of terms, choose a stable form, and check the result against the largest term or a direct short sum.

The case $r=1$ must be handled separately as $n a_1$ .

After the symbolic work is complete, check the result. Depending on the lesson, this may mean substituting into an original equation, comparing coordinates, examining a graph, checking units, testing an interval, or confirming that every branch of a periodic solution has been included.

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What commonly goes wrong

A common error is using exponent $n-1$ in the sum formula because the nth term uses $n-1$ .

The repair is to return to the definition and identify the first step where the invalid solution stops describing the original mathematical object. Later algebra cannot rescue a first step that changed the domain, orientation, branch, or meaning of the problem.

Textbook reading

Worked examples

Worked example 1

A ball rebounds to $70\%$ of its previous height. Starting from $10$ meters, find the total upward distance over the first $8$ rebounds.

Solution

Begin by identifying the mathematical object and the information that fixes it. Identify the first term, ratio, and number of terms, choose a stable form, and check the result against the largest term or a direct short sum. The relevant conditions are not optional bookkeeping: The case $r=1$ must be handled separately as $n a_1$ . Following that structure gives $\frac{10(0.7)(1-0.7^8)}{1-0.7}$ meters.

Why this works

The alternative form $\frac{a_1(r^n-1)}{r-1}$ is algebraically equivalent and may avoid nested negatives when $r>1$ . The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Transfer example

Problem

Derive formula by multiplying by $r$ and subtracting.

Worked development

Identify the first term, ratio, and number of terms, choose a stable form, and check the result against the largest term or a direct short sum. In this example, the first useful move is to make the defining structure visible rather than to search for a memorized answer. Multiplying S_n by $r$ shifts every term one position. Subtracting cancels the interior terms and leaves $a_1$ and $a_1 r^n,$ giving $S_n=\frac{a_1(1-r^n)}{1-r}$ for $r\ne 1$ . Then apply the conditions explicitly: The case $r=1$ must be handled separately as $n a_1$ . Finish by checking the result in a second representation and explaining what the result means.

Interpretation

Finite geometric sums model total rebounds, payments, repeated discounts, and digital scaling.

Reasoning example

Problem

Handle $r=1$ separately.

Worked development

Identify the first term, ratio, and number of terms, choose a stable form, and check the result against the largest term or a direct short sum. In this example, the first useful move is to make the defining structure visible rather than to search for a memorized answer. Multiplying S_n by $r$ shifts every term one position. Subtracting cancels the interior terms and leaves $a_1$ and $a_1 r^n,$ giving $S_n=\frac{a_1(1-r^n)}{1-r}$ for $r\ne 1$ . Then apply the conditions explicitly: The case $r=1$ must be handled separately as $n a_1$ . Finish by checking the result in a second representation and explaining what the result means.

Interpretation

Finite geometric sums model total rebounds, payments, repeated discounts, and digital scaling.

Worked example 4: quick check

Find $3+6+12+...+3\cdot 2^9$

Solution

Begin by identifying the mathematical object and the information that fixes it. Identify the first term, ratio, and number of terms, choose a stable form, and check the result against the largest term or a direct short sum. The relevant conditions are not optional bookkeeping: The case $r=1$ must be handled separately as $n a_1$ . Following that structure gives $3(2^10-1)=3069$ .

Why this works

The alternative form $\frac{a_1(r^n-1)}{r-1}$ is algebraically equivalent and may avoid nested negatives when $r>1$ . The calculation and the representation should agree, so a graph, diagram, table, or substitution check should support the same conclusion.

Textbook reading

Application and interpretation

Finite geometric sums model total rebounds, payments, repeated discounts, and digital scaling.

A contextual answer must include units, a meaningful domain, and the assumptions that make the model plausible. An exact mathematical relationship should not be diluted into a decimal unless a measurement or comparison requires it.

Practice