Induction for Geometric Series
A bank offers 5% annual compound interest. After $n$ years your total is linked to a geometric series. The sum formula $S_n = \dfrac{a(r^n - 1)}{r - 1}$ is used everywhere from finance to physics — but can you prove it holds for every positive integer $n$? Mathematical induction gives you a watertight proof in three steps. In this lesson you will carry out that proof and practise applying the formula.
The series $1 + 2 + 4 + 8 + \cdots$ doubles with each term. Without using a formula — estimate the sum of the first 10 terms. Write your reasoning below.
Every geometric series induction proof comes down to two moves: use the inductive hypothesis to rewrite the sum up to $k$, then add the next term $ar^k$ and simplify over the common denominator $(r-1)$.
The geometric series formula is:
$$S_n = a + ar + ar^2 + \cdots + ar^{n-1} = \frac{a(r^n - 1)}{r - 1}, \quad r \neq 1$$
In the inductive step we add $ar^k$ to both sides and show the numerator simplifies to $a(r^{k+1}-1)$.
Key facts
- $S_n = \dfrac{a(r^n - 1)}{r - 1}$ is the geometric series sum formula
- The formula requires $r \neq 1$
- The inductive step adds $ar^k$ and uses the common denominator $(r-1)$
Concepts
- Why the $ar^k$ terms cancel in the numerator during simplification
- Why the base case must explicitly evaluate both LHS and RHS
- The importance of stating $r \neq 1$ throughout the proof
Skills
- Write a complete induction proof for the geometric series formula
- Apply the formula to find the sum of a given geometric series
- Identify and avoid sign errors in the inductive step
A geometric series is a sum where each term is obtained by multiplying the previous term by a constant ratio $r$. The first term is $a$, so the series is:
Why does this work? Write $S = a + ar + \cdots + ar^{n-1}$. Multiply both sides by $r$: $rS = ar + ar^2 + \cdots + ar^n$. Subtract: $rS - S = ar^n - a$, so $S(r-1) = a(r^n - 1)$, giving $S = \dfrac{a(r^n-1)}{r-1}$.
Check your hook estimate: For $a=1$, $r=2$, $n=10$:
$S_{10} = \dfrac{1(2^{10}-1)}{2-1} = \dfrac{1024 - 1}{1} = \mathbf{1023}$
How close was your estimate?
Geometric series: $a+ar+ar^2+\cdots+ar^{n-1}=\frac{a(r^n-1)}{r-1}$, $r\neq1$. Derivation: $S-rS=a-ar^n \Rightarrow S(r-1)=a(r^n-1)$.
Pause — copy the geometric series formula $a+ar+\cdots+ar^{n-1}=\frac{a(r^n-1)}{r-1}$ and the algebraic derivation via $S-rS$ into your book.
Quick check: Which expression gives $S_5$ for the geometric series with $a = 3$ and $r = 2$?
We just saw that $a+ar+\cdots+ar^{n-1}=\frac{a(r^n-1)}{r-1}$ for $r\neq1$, derived algebraically via $S-rS=a-ar^n$. That raises a question: how does the inductive step handle adding $ar^k$ to $\frac{a(r^k-1)}{r-1}$ to produce $\frac{a(r^{k+1}-1)}{r-1}$? This card answers it → by expanding $ar^k(r-1)$ over the common denominator and cancelling the $ar^k$ terms.
We prove that $a + ar + ar^2 + \cdots + ar^{n-1} = \dfrac{a(r^n - 1)}{r - 1}$ for all positive integers $n$, where $r \neq 1$.
Step 1 — Base case ($n = 1$):
LHS $= a$
RHS $= \dfrac{a(r^1 - 1)}{r - 1} = \dfrac{a(r-1)}{r-1} = a$
LHS $=$ RHS. True for $n = 1$.
Step 2 — Inductive hypothesis:
Assume the statement is true for $n = k$ (where $k \geq 1$):
Step 3 — Inductive step ($n = k + 1$):
We need to prove:
LHS $= \underbrace{\dfrac{a(r^k - 1)}{r - 1}}_{\text{inductive hypothesis}} + ar^k$
$= \dfrac{a(r^k - 1) + ar^k(r - 1)}{r - 1}$
$= \dfrac{ar^k - a + ar^{k+1} - ar^k}{r - 1}$
$= \dfrac{ar^{k+1} - a}{r - 1}$
$= \dfrac{a(r^{k+1} - 1)}{r - 1}$ $=$ RHS
By the principle of mathematical induction, the formula holds for all $n \geq 1$ (where $r \neq 1$). $\blacksquare$
We prove that $a + ar + ar^2 + \cdots + ar^{n-1} = \dfrac{a(r^n - 1)}{r - 1}$ for all positive integers $n$, where $r \neq 1$.
Pause — copy the complete GP induction proof, highlighting the key cancellation $ar^k - ar^k = 0$ in the numerator into your book.
Did you get this? True or false: in the inductive step, $ar^k(r-1)$ expands to $ar^{k+1} - ar^k$.
Worked examples · 3 in a row, reveal as you go
Prove by induction that $a + ar + ar^2 + \cdots + ar^{n-1} = \dfrac{a(r^n - 1)}{r - 1}$ for all positive integers $n$, where $r \neq 1$.
Find the sum of the geometric series $4 + 12 + 36 + \cdots + 2916$.
Find the sum of the geometric series $2 + 6 + 18 + \cdots$ (10 terms).
Fill the gap: For the series $1 + 3 + 9 + \cdots$ (5 terms), $a = 1$, $r = 3$, so $S_5 = \dfrac{1(3^5 - 1)}{3 - 1} = \dfrac{242}{2} = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when $r = 1$ the sum of the first $n$ terms of a geometric series is $S_n = na$.
Activities · practice with the ideas
Write out the complete induction proof for the geometric series formula in your own words, labelling each step clearly.
Find the sum of the geometric series $5 + 10 + 20 + \cdots$ (8 terms).
A geometric series has first term 6, common ratio $\frac{1}{2}$, and last term $\frac{3}{16}$. Find the number of terms and the sum.
Explain in words why the $ar^k$ terms cancel in the numerator during the inductive step. No algebra required.
What goes wrong if you forget to state $r \neq 1$ in the proof? Give a specific numerical example to illustrate the problem.
Odd one out: Three of these geometric series sums are correct. Which one is NOT?
Earlier you estimated the sum of $1 + 2 + 4 + \cdots$ (10 terms).
The exact answer is $S_{10} = \dfrac{1(2^{10}-1)}{2-1} = \mathbf{1023}$. Did your intuition come close? Geometric series grow explosively — each doubling roughly doubles the running total. This is why the base case alone is never enough to establish a formula; induction is needed to capture the full pattern.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the sum of the geometric series $2 + 6 + 18 + \cdots$ (10 terms). (2 marks)
Q2. Prove by mathematical induction that $a + ar + ar^2 + \cdots + ar^{n-1} = \dfrac{a(r^n - 1)}{r - 1}$ for all positive integers $n$, where $r \neq 1$. (3 marks)
Q3. A geometric series has first term $a = 4$, common ratio $r = 3$, and sum $S_n = 4372$. Find $n$. (2 marks)
Comprehensive answers (click to reveal)
Activity answers: 2. $S_8 = \frac{5(2^8-1)}{1} = 5 \times 255 = 1275$ · 3. $6 \cdot (1/2)^{n-1} = 3/16 \Rightarrow (1/2)^{n-1} = 1/32 = (1/2)^5 \Rightarrow n = 6$; $S_6 = \frac{6(1-(1/2)^6)}{1-1/2} = \frac{6 \cdot 63/64}{1/2} = \frac{378/64}{1/2} = 378/32 = 189/16$. Alternatively using $r<1$ form: $S_6 = \frac{6(1-(0.5)^6)}{0.5} = 12(1-1/64)= 12 \times 63/64 = 756/64 = 189/16$.
Q1 (2 marks): $a=2$, $r=3$, $n=10$. $S_{10} = \dfrac{2(3^{10}-1)}{3-1} = \dfrac{2 \times 59048}{2} = \mathbf{59048}$ [1 for formula, 1 for answer].
Q2 (3 marks): Base case $n=1$: LHS $= a =$ RHS [1]. Hypothesis: assume $S_k = \frac{a(r^k-1)}{r-1}$ [implicit]. Inductive step: $S_{k+1} = \frac{a(r^k-1)}{r-1} + ar^k = \frac{a(r^k-1)+ar^k(r-1)}{r-1} = \frac{ar^{k+1}-a}{r-1} = \frac{a(r^{k+1}-1)}{r-1}$ [1 for correct algebra]. Conclusion by induction [1].
Q3 (2 marks): $4372 = \frac{4(3^n-1)}{2}$ [1] $\Rightarrow 3^n - 1 = 2186 \Rightarrow 3^n = 2187 = 3^7 \Rightarrow \mathbf{n = 7}$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering geometric series questions. Lighter alternative to the boss.
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