Mathematical-Reasoning · Unit 14 · Video 4 · Interactive Practice

Geometric Series by Induction: a(1 − rⁿ)/(1 − r)

IKey Formulas

Definition	Name	Proved by induction
$a_1 = a$ , $a_{n+1} = r \cdot a_n$	Geometric sequence	$a_n = a \cdot r^{n-1}$ for each $n \in \mathbb{N}$
$S_1 = a$ , $S_{n+1} = a + r \cdot S_n$	Geometric series	$S_n = a + ar + ar^2 + \cdots + ar^{n-1}$
same $S_n$ , with $r \ne 1$	Closed formula	$S_n = a \cdot \dfrac{1 - r^n}{1 - r}$ ; when $r = 1$ , $S_n = na$
$T_1 = 16$ , $T_{n+1} = 16 + \tfrac12 T_n$	The case $a = 16$ , $r = \tfrac12$	$T_n = 32\left(1 - (\tfrac12)^n\right)$ and $32 - T_n = 16 \cdot (\tfrac12)^{n-1} = b_n$

Key Insight: The inductive step for the closed formula rests on one cancellation in the numerator: $(1 - r) + r(1 - r^k) = 1 - r + r - r^{k+1} = 1 - r^{k+1}$ .

IIVisualization 1 — Why $T_n$ Approaches 32

The gap $32 - T_n = 32 \cdot (\tfrac12)^n$ is the halving term $b_n$ (from $b_1 = 16$ , $b_{n+1} = \tfrac12 b_n$ ), so it halves at every step.

n 3

💡 Because $b_n > 0$ for every $n \in \mathbb{N}$ , $T_n = 32 - b_n < 32$ for every $n$ , not only for the ten cases drawn.

IIIVisualization 2 — The Inductive Step, Line by Line

For $r \ne 1$ , every line of the inductive step's chain has the same value, $S_{k+1}$ ; at $r = 1$ the fractions are undefined.

Series k 3 Line 1

Line	Reason	With the numbers
$S_k = a \cdot \dfrac{1 - r^k}{1 - r}$	inductive hypothesis $P(k)$
$S_{k+1} = a + r \cdot S_k$	recurrence relation
$= a + r \cdot a \cdot \dfrac{1 - r^k}{1 - r}$	substitute $P(k)$ for $S_k$
$= a \cdot \dfrac{(1 - r) + r(1 - r^k)}{1 - r}$	write $a = a \cdot \frac{1 - r}{1 - r}$ , then factor out $a$
$= a \cdot \dfrac{1 - r^{k+1}}{1 - r}$	$1 - r + r - r^{k+1}$ : the $-r$ and $+r$ cancel, giving $P(k+1)$

IVQuiz Questions

Problem 1 · A Term of a Geometric Sequence

Given: a geometric sequence with $a_1 = 3$ and $a_{n+1} = 2 \cdot a_n$ for each $n \in \mathbb{N}$ . Find $a_6$ .

✅ Correct!

a_6 = 3 \cdot 2^{5} = 96

: five multiplications by

2

take

a_1

a_6

❌ One factor too many.

3 \cdot 2^6 = 192

uses the exponent

n

; the formula is

a_n = a \cdot r^{n-1}

, so

a_6

has exponent

5

. (

192

a_7

❌ That adds instead of multiplies.

3 + 5 \cdot 2 = 13

treats

2

as a common difference. In a geometric sequence each term is the previous one times

r

❌ The power applies to $r$ only.

(3 \cdot 2)^5 = 7776

raises

a

to the fifth power as well;

a \cdot r^{n-1}

multiplies

a

just once.

Show solution

Here $a = 3$ and $r = 2$ . By the first result, $a_n = a \cdot r^{n-1}$ for each $n \in \mathbb{N}$ , so

a_6 = 3 \cdot 2^{6-1} = 3 \cdot 32 = 96.

Check with the recurrence: $3, 6, 12, 24, 48, 96$ .

Problem 2 · The Case the Formula Excludes

Given: $S_1 = 1$ and $S_{n+1} = 1 + 1 \cdot S_n$ for each $n \in \mathbb{N}$ , the geometric series with $a = 1$ and $r = 1$ . Find $S_{10}$ .

✅ Correct! With

r = 1

every term

ar^{n-1}

equals

a

, so

S_{10} = 10 \cdot 1 = 10

❌ $\frac{0}{0}$ is not $0$ . Substituting

r = 1

into

\frac{1 - r^{10}}{1 - r}

gives

\frac{0}{0}

, which is undefined; that is exactly why the formula requires

r \ne 1

❌ That is a term, not the sum.

a_{10} = 1 \cdot 1^{9} = 1

is the tenth term of the sequence;

S_{10}

adds ten such terms.

❌ The series still exists. Only the closed formula fails at

r = 1

. The recurrence defines

S_{10}

perfectly well:

1, 2, 3, \dots, 10

Show solution

The closed formula $S_n = a \cdot \frac{1 - r^n}{1 - r}$ is proved only for $r \ne 1$ . For $r = 1$ use the sum form $S_n = a + ar + \cdots + ar^{n-1}$ : each of the $n$ terms equals $a$ , so $S_n = na$ .

S_{10} = 10 \cdot 1 = 10.

The recurrence agrees: $S_1 = 1$ , and each step adds $1$ .

Problem 3 · Read Off a and r, Then Sum

Given: $S_1 = 2$ and $S_{n+1} = 2 + 3 \cdot S_n$ for each $n \in \mathbb{N}$ . Find $S_5$ with the closed formula.

✅ Correct!

a = 2

r = 3

, and

S_5 = 2 \cdot \frac{1 - 243}{1 - 3} = 242

❌ That is the fifth term.

a_5 = 2 \cdot 3^4 = 162

is the last term of the sum;

S_5 = 2 + 6 + 18 + 54 + 162

❌ The factor $a$ is missing.

\frac{1 - 3^5}{1 - 3} = 121

is the sum for

a = 1

; multiply by

a = 2

❌ One term too many.

2 \cdot \frac{1 - 3^6}{1 - 3} = 728

S_6

; the exponent in

S_n

n

, so

S_5

uses

3^5

Show solution

Match $S_1 = a$ and $S_{n+1} = a + r \cdot S_n$ : the initial term is $a = 2$ and the ratio is $r = 3 \ne 1$ . Then

S_5 = 2 \cdot \frac{1 - 3^5}{1 - 3} = 2 \cdot \frac{1 - 243}{-2} = 2 \cdot 121 = 242.

Check with the recurrence: $2,\ 2 + 3 \cdot 2 = 8,\ 2 + 3 \cdot 8 = 26,\ 2 + 3 \cdot 26 = 80,\ 2 + 3 \cdot 80 = 242$ .

Problem 4 · The Gap for a Ratio of One Third

Given: the geometric series with $a = 1$ and $r = \tfrac13$ , so $S_n = 1 + \tfrac13 + \cdots + (\tfrac13)^{n-1}$ . Which expression equals $\tfrac32 - S_n$ for every $n \in \mathbb{N}$ ?

✅ Correct!

S_n = \tfrac32\left(1 - (\tfrac13)^n\right)

, so the gap is

\tfrac32 (\tfrac13)^n = \tfrac12 (\tfrac13)^{n-1}

, a geometric sequence itself, just as

32 - T_n = b_n

❌ A third of the gap. At

n = 1

\tfrac32 - S_1 = \tfrac32 - 1 = \tfrac12

, but this gives

\tfrac16

. The factor

\tfrac12

goes with

(\tfrac13)^{n-1}

, not

(\tfrac13)^n

❌ Three times the gap. At

n = 1

this gives

\tfrac32

, but

\tfrac32 - S_1 = \tfrac12

. The closed formula has

r^n

, not

r^{n-1}

❌ The factor $\frac{a}{1 - r}$ is missing. At

n = 1

this gives

\tfrac13

, but the gap is

\tfrac12

. Expanding

\tfrac32\left(1 - (\tfrac13)^n\right)

keeps the

\tfrac32

on both pieces.

Show solution

Since $r = \tfrac13 \ne 1$ , the closed formula applies, with $\frac{a}{1 - r} = \frac{1}{2/3} = \tfrac32$ :

S_n = 1 \cdot \frac{1 - (\tfrac13)^n}{1 - \tfrac13} = \tfrac32\left(1 - (\tfrac13)^n\right) = \tfrac32 - \tfrac32 \cdot (\tfrac13)^n.

So $\tfrac32 - S_n = \tfrac32 \cdot (\tfrac13)^n$ . Check at $n = 2$ : $S_2 = \tfrac43$ , and $\tfrac32 - \tfrac43 = \tfrac16 = \tfrac32 \cdot \tfrac19$ ✓.

Solved: 0 / 4

Geometric Series by Induction: a(1 − rⁿ)/(1 − r)

IKey Formulas

IIVisualization 1 — Why TnT_nTn​ Approaches 32

IIIVisualization 2 — The Inductive Step, Line by Line

IVQuiz Questions

IIVisualization 1 — Why $T_n$ Approaches 32