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Module 5 · L4 of 20 ~45 min ⚡ +95 XP available

Induction: Arithmetic Series

You know arithmetic sequences cold, first term $a$, common difference $d$, $n$-th term $a+(n-1)d$. But how do you prove that the sum formula $S_n = \dfrac{n}{2}[2a+(n-1)d]$ works for every positive integer $n$? In this lesson you will run the full induction proof, master the tricky algebra in the inductive step, and apply the formula to real sums.

Today's hook, Consider the sequence $5, 8, 11, 14, 17, \ldots$ (first term 5, common difference 3). Without any formula, estimate the sum of the first 10 terms. Jot your guess, you'll compute the exact answer using the induction-proven formula in the activities.

0/5QUESTS

Recall, your gut answer first

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For the sequence $5, 8, 11, 14, 17, \ldots$, estimate the sum of the first 10 terms without using a formula. Describe your reasoning or strategy.

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Think first

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The arithmetic series formula $S_n = \dfrac{n}{2}[2a + (n-1)d]$ generalises Lesson 3's result to any first term $a$ and common difference $d$. When $a = 1$ and $d = 1$ it reduces to $\dfrac{n(n+1)}{2}$.

The inductive step is more algebra-intensive here because we have two parameters ($a$ and $d$) to carry through. The strategy remains the same: add the $(k+1)$-th term to the hypothesis and manipulate until the result matches the target formula with $k+1$ substituted for $n$.

$(k+1)$-th term of AP $= a + kd$

Target: $\dfrac{k+1}{2}[2a + kd]$

$$S_{k+1} = S_k + [a + kd] = \frac{k+1}{2}[2a+kd]$$

The $(k+1)$-th term

In a general AP, the $(k+1)$-th term is $a + kd$, substitute $n = k+1$ into $a + (n-1)d$.

Common denominator

Combine the hypothesis (with denominator 2) and the next term over a common denominator of 2 before factorising.

Factorise $(k+1)$

After combining, factor $(k+1)$ from the numerator to reveal $\dfrac{k+1}{2}[2a + kd]$.

What you'll master

Know

Key facts

$S_n = \dfrac{n}{2}[2a + (n-1)d]$ is the arithmetic series formula
The $(k+1)$-th term of an AP is $a + kd$
Equivalent form: $S_n = \dfrac{n}{2}(a + l)$ where $l$ is the last term

Understand

Concepts

Why the inductive step requires adding $a + kd$ (not $a + (k+1)d$) to the hypothesis
How combining over a common denominator of 2 leads to factorisation of $(k+1)$
Why $S_n = \dfrac{n(n+1)}{2}$ is a special case with $a = d = 1$

Can do

Skills

Write a complete induction proof for the arithmetic series formula
Apply the formula to find the number of terms and sum of any given AP
Correctly expand and collect $k(k-1)d$ and $2kd$ terms in the inductive step

Key terms

Arithmetic seriesThe sum of the terms of an arithmetic sequence: $a + (a+d) + (a+2d) + \cdots$

First term ($a$)The value of the first term in the sequence. Also called $T_1$ or $u_1$.

Common difference ($d$)The constant amount added to each term to get the next. Can be positive, negative, or zero.

General termThe $n$-th term of the AP is $T_n = a + (n-1)d$. The $(k+1)$-th term is $a + kd$.

$S_n$The sum of the first $n$ terms: $S_n = \dfrac{n}{2}[2a + (n-1)d]$, equivalently $\dfrac{n}{2}(a+l)$ where $l = a + (n-1)d$.

Inductive step algebraThe process of combining $\dfrac{k}{2}[2a+(k-1)d] + (a+kd)$ over a denominator of 2 and factorising $(k+1)$ to reach the target.

The arithmetic series formula

core concept

An arithmetic series is the sum of an arithmetic sequence. If the first term is $a$ and the common difference is $d$, the sum of the first $n$ terms is:

$$S_n = a + (a+d) + (a+2d) + \cdots + [a+(n-1)d] = \frac{n}{2}[2a + (n-1)d]$$

This can also be written as $S_n = \dfrac{n}{2}(a + l)$ where $l = a + (n-1)d$ is the last term.

Quick check: For $a = 1$, $d = 1$: $S_n = \dfrac{n}{2}[2 + (n-1)] = \dfrac{n(n+1)}{2}$, this is Lesson 3's formula, confirming consistency.

Finding $n$ given the last term. If you know $a$, $d$, and the last term $l$, solve $a + (n-1)d = l$ for $n$: $n = \dfrac{l - a}{d} + 1$. Example: last term $32$, $a = 5$, $d = 3$: $n = \dfrac{32-5}{3} + 1 = 9 + 1 = 10$.

An arithmetic series: $S_n = \frac{n}{2}[2a+(n-1)d]$ where $a$ is the first term and $d$ is the common difference. Equivalently $S_n=\frac{n}{2}(a+l)$ where $l=a+(n-1)d$ is the last term.

Pause, copy the arithmetic series formula $S_n=\frac{n}{2}[2a+(n-1)d]$ and note that the $(k+1)$-th term is $a+kd$ into your book.

Quick check: What is the $(k+1)$-th term of an arithmetic sequence with first term $a$ and common difference $d$?

Proving the formula by induction

core concept

We just saw that $S_n=\frac{n}{2}[2a+(n-1)d]$ and that the $(k+1)$-th term is $a+kd$. That raises a question: how do you combine the hypothesis $S_k=\frac{k}{2}[2a+(k-1)d]$ with $a+kd$ to reach $\frac{k+1}{2}[2a+kd]$? This card answers it → combine over denominator 2, expand, then use $k(k-1)+2k=k(k+1)$ and factorise $(k+1)$.

We prove: $a + (a+d) + \cdots + [a+(n-1)d] = \dfrac{n}{2}[2a+(n-1)d]$ for all positive integers $n$.

Step 1, Base case ($n = 1$):

LHS $= a$. RHS $= \dfrac{1}{2}[2a + 0] = a$. LHS = RHS. True for $n = 1$. ✓

Step 2, Inductive hypothesis:

Assume true for $n = k$: $a + (a+d) + \cdots + [a+(k-1)d] = \dfrac{k}{2}[2a+(k-1)d]$

Step 3, Inductive step ($n = k+1$):

We need to show: $a + \cdots + [a+(k-1)d] + (a+kd) = \dfrac{k+1}{2}[2a+kd]$

LHS $= \dfrac{k}{2}[2a+(k-1)d] + (a+kd)$

$= \dfrac{k[2a+(k-1)d] + 2(a+kd)}{2}$

$= \dfrac{2ak + k(k-1)d + 2a + 2kd}{2}$

$= \dfrac{2a(k+1) + d[k(k-1) + 2k]}{2}$

$= \dfrac{2a(k+1) + d[k^2 - k + 2k]}{2}$

$= \dfrac{2a(k+1) + dk(k+1)}{2}$

$= \dfrac{(k+1)[2a + kd]}{2} = \dfrac{k+1}{2}[2a+kd]$ = RHS. ✓

Conclusion: By the principle of mathematical induction, the result is true for all positive integers $n \geq 1$. $\blacksquare$

Key algebra move. The step $k(k-1)+2k = k^2 - k + 2k = k^2 + k = k(k+1)$ is the hinge of the whole proof. Practise it until it is automatic.

We prove: $a + (a+d) + \cdots + [a+(n-1)d] = \dfrac{n}{2}[2a+(n-1)d]$ for all positive integers $n$.

Pause, copy the complete AP sum induction proof, highlighting the key simplification $k(k-1)+2k=k(k+1)$ into your book.

Did you get this? True or false: in the inductive step for the AP sum, the key simplification is $k(k-1) + 2k = k(k+1)$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · FULL INDUCTION PROOF

Prove by mathematical induction that $a+(a+d)+(a+2d)+\cdots+[a+(n-1)d] = \dfrac{n}{2}[2a+(n-1)d]$ for all positive integers $n$.

Base case ($n=1$): LHS $= a$; RHS $= \dfrac{1}{2}[2a+0] = a$. True for $n=1$. ✓

Substitute $n=1$. The $d$ term vanishes: $(n-1)d = 0$. Verify LHS = RHS explicitly.

PROBLEM 2 · APPLYING THE FORMULA

Find the sum of the series $5 + 8 + 11 + \cdots + 32$.

Identify: $a = 5$, $d = 3$, last term $l = 32$. Find $n$: $5 + 3(n-1) = 32 \Rightarrow 3(n-1) = 27 \Rightarrow n = 10$.

Before using the sum formula you must know $n$. Solve $a+(n-1)d = l$ for $n$.

PROBLEM 3 · EXPAND AND COLLECT

In the inductive step for the AP sum formula, show the full expansion that simplifies $\dfrac{k}{2}[2a+(k-1)d] + (a+kd)$ to $\dfrac{k+1}{2}[2a+kd]$.

$\dfrac{k[2a+(k-1)d] + 2(a+kd)}{2} = \dfrac{2ak + k(k-1)d + 2a + 2kd}{2}$

Place the next term over the common denominator 2, then distribute $k$ into the bracket and $2$ into $(a+kd)$.

Fill the gap: $k(k-1) + 2k = k^2 + $$= k(k+1)$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using the wrong next term: $a + (k+1)d$ instead of $a + kd$

The $(k+1)$-th term of an AP is obtained by substituting $n = k+1$ into $a+(n-1)d$, giving $a+kd$, NOT $a+(k+1)d$. Getting this wrong breaks the whole inductive step. Always substitute carefully and write out the term explicitly before adding it to the hypothesis.

Trap 02

Sign errors when expanding $k(k-1)d$

When you expand $k(k-1)d$, the result is $k^2 d - kd$, not $k^2 d + kd$. Then adding $2kd$ gives $k^2 d + kd = kd(k+1)$. A sign error here means the factorisation will not work. Write each step on a new line and check signs carefully.

Trap 03

Not simplifying to the exact target form

The inductive step must produce exactly $\dfrac{k+1}{2}[2a + kd]$. Some students stop at $\dfrac{(k+1)(2a+kd)}{2}$ and do not rewrite it in the bracket form. Show that your expression matches the formula with $n = k+1$ substituted in by stating it explicitly.

Did you get this? True or false: the $(k+1)$-th term of an AP with first term $a$ and common difference $d$ is $a + (k+1)d$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Find the sum of the series $5 + 8 + 11 + \cdots$ (first 10 terms) using $S_n = \dfrac{n}{2}[2a+(n-1)d]$ with $a = 5$, $d = 3$, $n = 10$. Check your earlier estimate.

Find the sum of $3 + 7 + 11 + \cdots$ (20 terms).

Write out the base case for the AP sum formula proof. Show LHS and RHS separately for $n = 1$.

Simplify $k(k-1) + 2k$ step by step and state the factored form.

An AP has first term $a = 2$, common difference $d = 5$, and last term $47$. Find $n$ and then compute $S_n$.

Odd one out: Three of these statements about the AP sum formula are correct. Which one is NOT?

Revisit your thinking

Earlier you estimated the sum $5 + 8 + 11 + \cdots$ (10 terms). The exact answer is $S_{10} = \dfrac{10}{2}[2(5)+9(3)] = 5[10+27] = 5 \times 37 = \mathbf{185}$.

The formula packs a lot of algebra, but every step in the induction proof followed a clear recipe: add the next term, combine, factorise. Did your estimate overshoot or undershoot? Why do you think that is?

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Multiple choice

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Short answer

ApplyBand 32 marks

Q1. Find the sum of $3 + 7 + 11 + \cdots$ (20 terms). (2 marks)

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ApplyBand 43 marks

Q2. Prove by mathematical induction that $a + (a+d) + (a+2d) + \cdots + [a+(n-1)d] = \dfrac{n}{2}[2a+(n-1)d]$ for all positive integers $n$. (3 marks)

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AnalyseBand 52 marks

Q3. In the inductive step, a student writes the next term as $a + (k+1)d$ instead of $a + kd$. Identify the error and explain its consequence. (2 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $S_{10} = \dfrac{10}{2}[10+27] = 5 \times 37 = \mathbf{185}$.

2. $a=3$, $d=4$, $n=20$: $S_{20} = \dfrac{20}{2}[6+19\times4] = 10[6+76] = 10 \times 82 = \mathbf{820}$.

3. LHS $= a$; RHS $= \dfrac{1}{2}[2a + 0] = a$. LHS = RHS. True for $n = 1$.

4. $k(k-1)+2k = k^2 - k + 2k = k^2 + k = k(k+1)$.

5. $2 + 5(n-1) = 47 \Rightarrow 5(n-1) = 45 \Rightarrow n = 10$. $S_{10} = \dfrac{10}{2}[4 + 9\times5] = 5 \times 49 = \mathbf{245}$.

Q1 (2 marks): $a=3$, $d=4$, $n=20$ [1]. $S_{20} = \dfrac{20}{2}[2(3)+19(4)] = 10[6+76] = 10 \times 82 = \mathbf{820}$ [1].

Q2 (3 marks): Base case: $n=1$, LHS $= a$, RHS $= \dfrac{1}{2}[2a] = a$. True [1]. Hypothesis: assume $S_k = \dfrac{k}{2}[2a+(k-1)d]$. Inductive step: $S_{k+1} = \dfrac{k}{2}[2a+(k-1)d]+(a+kd) = \dfrac{(k+1)}{2}[2a+kd]$ = formula with $n=k+1$ [1]. By induction, true for all $n \geq 1$ [1].

Q3 (2 marks): The $(k+1)$-th term is $a + kd$ (from $a+(n-1)d$ with $n=k+1$), not $a+(k+1)d$ which would be the $(k+2)$-th term [1]. Consequence: adding the wrong term means the algebraic expression cannot be factorised to match the target formula $\dfrac{k+1}{2}[2a+kd]$, so the inductive step fails [1].

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Module overview · Extension 1 · Checkpoint 1