Induction proofs of simple sums and arithmetic/geometric-series results | Year 11

01 / LEARN

Understand the idea.

Fixed parameters, changing index

In a parameterised sum, quantities such as $a, d, r$ remain fixed throughout induction. Only the number of terms changes. After $k$ arithmetic terms, add $a + k d$ ; after $k$ geometric terms beginning at $a$ , add $a r^{k}$ .

Write the required $k + 1$ formula so the target is clear. Begin the calculation with the actual next sum, replace its first $k$ terms using the hypothesis, then collect and factor until the target form appears.

Restrictions belong in the statement

State restrictions before beginning. For the general finite geometric fraction, $r \neq 1$ is essential because the denominator is $1 - r$ . The case $r = 1$ consists of $n$ equal terms and has sum $n a$ . Finite sums otherwise need no convergence restriction.

A complete proof includes the first case, a hypothesis at arbitrary $k$ , a justified implication to $k + 1$ , and the final integer domain. Show the key algebra rather than claiming that the next case follows merely because several small values work.

\sum_{j = 1}^{n} [a + (j - 1) d] = \frac{n}{2} [2 a + (n - 1) d]

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A general arithmetic sum

Prove the general arithmetic-sum formula for fixed real $a, d$ .

Worked solution

1
Verify the formula for one term.
$S_{1} = a = \frac{1}{2} [2 a + (1 - 1) d]$
The formula must hold for every fixed real choice of a and d at the base index.
2
State the induction hypothesis.
$S_{k} = \frac{k}{2} [2 a + (k - 1) d], k \geq 1$
Assume the formula for a sum of k terms, keeping a and d fixed.
3
Add term k plus one.
$\begin{aligned} S_{k + 1} & = S_{k} + (a + k d) \\ = \frac{2 a k + k (k - 1) d + 2 a + 2 k d}{2} \end{aligned}$
The arithmetic term after $a + (k - 1) d$ is $a + k d$ .
4
Collect and factor to the next-case form.
$\begin{aligned} S_{k + 1} & = \frac{2 a (k + 1) + k (k + 1) d}{2} \\ = \frac{k + 1}{2} (2 a + k d) \end{aligned}$
The d terms combine using $k (k - 1) + 2 k = k (k + 1)$ .
5
Conclude for positive integer counts.
$S_{n} = \frac{n}{2} [2 a + (n - 1) d] (n \in ℤ, n \geq 1)$
The derived expression is the stated formula with n replaced by k plus one.

Answer

$S_{n} = n [2 a + (n - 1) d] / 2$ for integers $n \geq 1$ .

EXAMPLE 02

A finite geometric identity

Prove $1 + 2 + \dots + 2^{n - 1} = 2^{n} - 1$ .

Worked solution

1
Verify the base case.
$n = 1 : 1 = 2^{1} - 1$
The first sum has only the term $2^{0} = 1$ .
2
Assume the k-term identity.
$S_{k} = 1 + 2 + \dots + 2^{k - 1} = 2^{k} - 1$
This hypothesis concerns terms with exponents zero through k minus one.
3
Append the next power.
$\begin{aligned} S_{k + 1} & = S_{k} + 2^{k} \\ = (2^{k} - 1) + 2^{k} \\ = 2^{k + 1} - 1 \end{aligned}$
The new term has exponent k, and two copies of $2^{k}$ make $2^{k + 1}$ .
4
Conclude the identity by induction.
$1 + 2 + \dots + 2^{n - 1} = 2^{n} - 1 (n \in ℤ, n \geq 1)$
The true base case and the general implication cover every positive integer.

Answer

True for integers $n \geq 1$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

What is added after the $k$ case in $3 + 7 + \dots + (4 n - 1)$ ?

Show answer & reasoning

$4 k + 3$ .

The k-term sum ends at $4 k - 1$ . Replace the index by $k + 1$ to obtain the next term: $4 (k + 1) - 1 = 4 k + 3$ .

Keep practising

6 worksheet questions · 19 marks

Open worksheet Worked solutions

Read the questions online

Prove $3 + 7 + \dots + (4 n - 1) = n (2 n + 1)$ .
4 marks
Prove $1 + 3 + \dots + 3^{n - 1} = \frac{3^{n} - 1}{2}$ .
4 marks
State the next term and required restriction in a proof of $a + a r + \dots + a r^{n - 1} = \frac{a (1 - r^{n})}{1 - r}$ .
2 marks
Simplify $\frac{a (1 - r^{k})}{1 - r} + a r^{k}$ to the next-case formula.
3 marks
Use an arithmetic sum to find $5 + 8 + 11 + \dots + 62$ .
2 marks
Prove $2 + 6 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$ .
4 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Prove $3 + 7 + \dots + (4 n - 1) = n (2 n + 1)$ .
4 marks
Base: $3 = 1 (3)$ . Assume $S_{k} = k (2 k + 1)$ .
Add $4 k + 3$ to get $2 k^{2} + 5 k + 3 = (k + 1) (2 k + 3)$ ; conclude.
True for integers $n \geq 1$ .
Prove $1 + 3 + \dots + 3^{n - 1} = \frac{3^{n} - 1}{2}$ .
4 marks
Base: $1 = \frac{3 - 1}{2}$ . Assume $S_{k} = \frac{3^{k} - 1}{2}$ .
Add $3^{k}$ to get $\frac{3^{k + 1} - 1}{2}$ ; conclude.
True for integers $n \geq 1$ .
State the next term and required restriction in a proof of $a + a r + \dots + a r^{n - 1} = \frac{a (1 - r^{n})}{1 - r}$ .
2 marks
The denominator must not be zero.
Next term $a r^{k}$ ; restriction $r \neq 1$ .
Simplify $\frac{a (1 - r^{k})}{1 - r} + a r^{k}$ to the next-case formula.
3 marks
The numerator becomes $a [1 - r^{k} + r^{k} (1 - r)] = a (1 - r^{k + 1})$ .
$\frac{a (1 - r^{k + 1})}{1 - r}$ .
Use an arithmetic sum to find $5 + 8 + 11 + \dots + 62$ .
2 marks
$62 = 5 + 3 (n - 1)$ gives $n = 20$ ; $\frac{20 (5 + 62)}{2} = 670$ .
$670$ .
Prove $2 + 6 + \dots + 2 \cdot 3^{n - 1} = 3^{n} - 1$ .
4 marks
Base: $2 = 3 - 1$ . Assume $S_{k} = 3^{k} - 1$ .
Add $2 \cdot 3^{k}$ to obtain $3^{k + 1} - 1$ ; conclude.
True for integers $n \geq 1$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics sequence and series content with introductory induction, regrouped into an original school-sized chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Fixed parameters, changing index

Restrictions belong in the statement

See the method in action.

A general arithmetic sum

Verify the formula for one term.

State the induction hypothesis.

Add term k plus one.

Collect and factor to the next-case form.

Conclude for positive integer counts.

A finite geometric identity

Verify the base case.

Assume the k-term identity.

Append the next power.

Conclude the identity by induction.

Try it yourself.

Check your understanding

Keep practising