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

01 / LEARN

The idea, step by step

Fixed parameters, changing index

In a formula containing $a, d$ or $r$ , those parameters remain fixed as $n$ changes. The induction step adds one next term: $a + k d$ for an arithmetic sum or $a r^{k}$ for a geometric sum.

Writing the required $k + 1$ form shows the target, not a fact you can assume. Begin from the actual next sum and use the $k$ hypothesis to derive the target.

Restrictions belong in the statement

The geometric fraction $a (1 - r^{n}) / (1 - r)$ requires $r \neq 1$ . When $r = 1$ the sum is $n a$ and must be considered separately.

A complete proof verifies the base, states the hypothesis, shows the cancellation or factorisation in the implication step and concludes the stated integer domain.

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

A common trap

After the first $k$ geometric terms, add $a r^{k + 1}$ .

Instead: The first $k$ terms end at $a r^{k - 1}$ , so add $a r^{k}$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

A general arithmetic sum

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

Base: $S_{1} = a$ .
Assume the formula for $k$ . Add $a + k d$ to give $[2 a k + k (k - 1) d + 2 a + 2 k d] / 2$ .
Factor to $(k + 1) (2 a + k d) / 2$ ; conclude by induction.

S_{n} = n [2 a + (n - 1) d] / 2

for integers

n \geq 1

Example 2

A finite geometric identity

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

Base: $1 = 2 - 1$ .
Assume $S_{k} = 2^{k} - 1$ .
Add $2^{k}$ : $S_{k + 1} = 2 (2^{k}) - 1 = 2^{k + 1} - 1$ . Conclude.

True for integers

n \geq 1

Check your understanding

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

Show answer

$4 k + 3$ .

The next term is $4 (k + 1) - 1$ .

03 / WORKSHEET

Your turn

6 questions · 19 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

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

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