What is no concluding statement?

Always finish with the sentence that, by induction, the result holds for all positive integers n.

Evaluate _r=1^n r^3 in factorised form. [1 mark]

SQA AH style: summation-style practice: Find _r=1^n (r^2 + 2r) in terms of n, simplified.

Split the sum: r^2 + 2sum r (1 mark). Use the formulae: n(n + 1)(2n + 1)6 + 2·n(n + 1)2 (1 mark). = n(n + 1)(2n + 1)6 + n(n + 1) (1 mark). Factor n(n + 1)6: n(n + 1)6[(2n + 1) + 6] = n(n + 1)(2n + 7)6 (1 mark). Markers reward splitting, the standard formulae, and a fully factorised simplification.

ScotlandMathsSyllabus dot point

How do you sum a series using standard formulae and prove a result for all positive integers by induction?

Apply the standard summation formulae for the sum of the first n natural numbers, their squares and their cubes, use sigma notation, and prove statements about series, divisibility and inequalities for all positive integers by mathematical induction.

A focused answer to the SQA Advanced Higher Mathematics summation and proof by induction content, covering sigma notation, the standard formulae for the sum of the first n natural numbers, squares and cubes, and the structure of a proof by mathematical induction applied to series, divisibility and inequalities.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Sigma notation and the standard sums
The structure of a proof by induction
Induction for series
Induction for divisibility
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What this dot point is asking

The SQA wants you to sum series using the standard sigma formulae and to prove results for all positive integers by mathematical induction. Induction questions cover series, divisibility and inequalities, and a full, correctly structured proof is what earns the marks.

Sigma notation and the standard sums

Sigma notation compresses a sum. Because summation is linear, you can split a sum across addition and pull out constant factors, then apply the standard formulae term by term.

Summing the first

n

odd numbers using sigma linearity

Evaluate $\displaystyle\sum_{r=1}^{n} (2r - 1)$ .

Step 1: Split using linearity

Because summation is linear, we can break the sum into two separate sums and pull out constant factors. This converts a sum we do not know directly into sums we do:

\sum_{r=1}^{n} (2r - 1) = 2\sum_{r=1}^{n} r - \sum_{r=1}^{n} 1.

Step 2: Apply the standard sums

Substitute the known formulae. The second sum, adding the constant $1$ for $n$ values, equals $n$ :

= 2\cdot\frac{n(n + 1)}{2} - n = n(n + 1) - n.

Step 3: Simplify

= n^2 + n - n = n^2.

Final answer: $\displaystyle\sum_{r=1}^{n} (2r - 1) = n^2$ . The first $n$ odd numbers always sum to a perfect square.

The structure of a proof by induction

Mathematical induction proves a statement $P(n)$ for every positive integer. Think of it as toppling dominoes: knock over the first, and show each one knocks over the next.

Induction for series

For a series result, the inductive step adds the $(k + 1)$ th term to the assumed sum for $k$ and shows the total matches the formula with $n = k + 1$ .

Proving the sum-of-squares formula by induction

Prove $\displaystyle\sum_{r=1}^{n} r^2 = \dfrac{n(n + 1)(2n + 1)}{6}$ by induction.

Step 1: Base case

Verify the formula works for the smallest case $n = 1$ . The left side is just $1^2 = 1$ ; the right side is:

\frac{1 \cdot 2 \cdot 3}{6} = 1.

Both sides agree, so the statement is true for $n = 1$ .

Step 2: Assumption

Assume the statement is true for some positive integer $k$ , that is:

\sum_{r=1}^{k} r^2 = \frac{k(k + 1)(2k + 1)}{6}.

Step 3: Inductive step

We must show the statement also holds for $k + 1$ . Add the $(k+1)$ th term to both sides of the assumed result, then factor out $(k+1)$ :

\sum_{r=1}^{k+1} r^2 = \frac{k(k + 1)(2k + 1)}{6} + (k + 1)^2 = \frac{(k + 1)\left[k(2k + 1) + 6(k + 1)\right]}{6}

= \frac{(k + 1)(2k^2 + 7k + 6)}{6} = \frac{(k + 1)(k + 2)(2k + 3)}{6},

which is exactly the formula with $n$ replaced by $k + 1$ .

Step 4: Conclude

The statement is true for $n = 1$ , and truth for any $k$ forces truth for $k + 1$ , so by mathematical induction it holds for all positive integers $n$ .

Induction for divisibility

For a divisibility result such as " $f(n)$ is divisible by $m$ ", write $f(k + 1)$ in terms of $f(k)$ plus a multiple of $m$ , so the assumption that $m \mid f(k)$ forces $m \mid f(k + 1)$ .

Proving divisibility by induction

Prove that $7^{n} - 1$ is divisible by $6$ for all positive integers $n$ .

Step 1: Base case

Check $n = 1$ : $7^1 - 1 = 6$ , and $6$ is divisible by $6$ . The statement holds for $n = 1$ .

Step 2: Assumption

Assume the statement holds for some positive integer $k$ , meaning $7^k - 1$ is divisible by $6$ . We write this as $7^k - 1 = 6m$ for some integer $m$ , which rearranges to $7^k = 6m + 1$ .

Step 3: Inductive step

We must show $7^{k+1} - 1$ is divisible by $6$ . Rewrite $7^{k+1}$ using the assumption:

7^{k+1} - 1 = 7 \cdot 7^k - 1 = 7(6m + 1) - 1 = 42m + 7 - 1 = 42m + 6 = 6(7m + 1).

Since $7m + 1$ is an integer, $6 \mid 7^{k+1} - 1$ .

Step 4: Conclude

The statement is true for $n = 1$ , and truth for any $k$ forces truth for $k + 1$ , so by mathematical induction $6 \mid 7^n - 1$ for all positive integers $n$ .

Try this

Q1. Evaluate $\displaystyle\sum_{r=1}^{n} r^3$ in factorised form. [1 mark]

Cue. $\left[\dfrac{n(n + 1)}{2}\right]^2$ .

Q2. State the four parts of a proof by induction. [2 marks]

Cue. Base case, assumption for $k$ , inductive step to $k + 1$ , concluding statement.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: summation4 marksFind

\displaystyle\sum_{r=1}^{n} (r^2 + 2r)

in terms of

n

, simplified.

Show worked answer →

Split the sum: $\displaystyle\sum r^2 + 2\sum r$ (1 mark).

Use the formulae: $\dfrac{n(n + 1)(2n + 1)}{6} + 2\cdot\dfrac{n(n + 1)}{2}$ (1 mark).

$= \dfrac{n(n + 1)(2n + 1)}{6} + n(n + 1)$ (1 mark).

Factor $\dfrac{n(n + 1)}{6}$ : $\dfrac{n(n + 1)}{6}\left[(2n + 1) + 6\right] = \dfrac{n(n + 1)(2n + 7)}{6}$ (1 mark). Markers reward splitting, the standard formulae, and a fully factorised simplification.

AH style: induction5 marksProve by induction that

\displaystyle\sum_{r=1}^{n} r = \dfrac{n(n + 1)}{2}

for all positive integers

n

Show worked answer →

Base case $n = 1$ : left side $= 1$ , right side $= \dfrac{1\times 2}{2} = 1$ , so true for $n = 1$ (1 mark).

Assume true for $n = k$ : $\displaystyle\sum_{r=1}^{k} r = \dfrac{k(k + 1)}{2}$ (1 mark).

Inductive step: $\displaystyle\sum_{r=1}^{k+1} r = \dfrac{k(k + 1)}{2} + (k + 1) = (k + 1)\left(\dfrac{k}{2} + 1\right) = \dfrac{(k + 1)(k + 2)}{2}$ (2 marks).

This is the formula with $n = k + 1$ . Since true for $n = 1$ and truth for $k$ implies truth for $k + 1$ , it holds for all positive integers by induction (1 mark). Markers reward the base case, the assumption, the algebra of the step, and the concluding statement.

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)