Mathematical induction for series identities

Syllabus dot point

“Prove identities for sums of series using the principle of mathematical induction”

HSCMaths Extension 1Proof (ME-P1)12 min readUpdated 20 Jun 2026

Quick answer

To prove a series identity by induction, verify the base case at $n = 1$ , assume the identity at $n = k$ , derive it at $n = k + 1$ by splitting off the last term and substituting the hypothesis, then conclude by the principle of mathematical induction. The whole method is the domino chain: one push plus "each topples the next" makes them all fall.

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What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

NESA wants you to use mathematical induction to prove that a formula for a finite sum is correct for every positive integer $n$ , not just for the few values you could check by hand. A series identity is a claim of the form "this running total equals this closed expression", and induction is the natural tool because the running total at $n = k + 1$ is built directly out of the running total at $n = k$ . The proof always has the same four named parts, and the marks are spread across all four, so reproducing the structure exactly is half the task.

The answer

The principle of mathematical induction

The principle of mathematical induction lets you prove infinitely many statements at once from two finite checks. Let $P(n)$ be a statement about a positive integer $n$ . If

$P(1)$ is true (the base case), and
for every positive integer $k$ , the implication $P(k) \implies P(k + 1)$ holds (the inductive step),

then $P(n)$ is true for all integers $n \ge 1$ .

The picture to keep in your head is a row of dominoes. The base case knocks the first one over; the inductive step guarantees that whichever domino falls knocks over the next; together they topple the whole infinite chain.

Two checks, not infinitely many: that is the whole power of the method. Notice that the base case may start at $n = 0$ or some other integer rather than $n = 1$ if that is what the statement asserts; start the chain at whatever the smallest claimed value of $n$ is.

Key fact

A proof by induction needs exactly two things to be watertight: the base case $P(1)$ verified by direct substitution, and the inductive step that derives $P(k + 1)$ from the assumed $P(k)$ . Miss either and the chain breaks. The conclusion sentence ("by the principle of mathematical induction...") then states what the two parts together prove.

The standard four-part structure

The four-part structure is the template every series proof must follow, and you should write each part with its own heading or label so the marker can tick them off.

Part 1: Base case: Verify $P(1)$ by substituting $n = 1$ into both sides separately and checking they are equal. Compute the two sides independently; never assume they match.
Part 2: Inductive hypothesis: Assume $P(k)$ holds for some positive integer $k$ , and write the assumed identity out in full. This line is the fact you are allowed to use later, so make it explicit.
Part 3: Inductive step: Prove $P(k + 1)$ using the hypothesis. For a series, the move is mechanical: write the $(k + 1)$ -term sum as the $k$ -term sum plus the new last term, replace the $k$ -term sum by the formula the hypothesis gives, then simplify to the closed form with $n = k + 1$ substituted.
Part 4: Conclusion: State that since the base case holds and $P(k) \implies P(k + 1)$ , by the principle of mathematical induction $P(n)$ holds for all positive integers $n$ .

Common series formulas worth memorising

The four standard sum formulas come up constantly, both as the thing you must prove and as a result you can quote elsewhere:

\sum_{i = 1}^{n} i = \frac{n(n + 1)}{2},

\sum_{i = 1}^{n} i^2 = \frac{n(n + 1)(2 n + 1)}{6},

\sum_{i = 1}^{n} i^3 = \left( \frac{n(n + 1)}{2} \right)^2,

\sum_{i = 0}^{n - 1} a r^i = a \cdot \frac{r^n - 1}{r - 1} \quad (r \neq 1).

You should be able to prove each of these on demand. A neat sanity check: the sum of cubes equals the square of the sum of integers, so $\sum i^3 = \left( \sum i \right)^2$ .

The split-off-the-last-term technique

The technique that drives the inductive step is always the same: peel off the final term so the rest matches the hypothesis. In symbols,

\sum_{i = 1}^{k + 1} a_i = \underbrace{\sum_{i = 1}^{k} a_i}_{\text{use the hypothesis here}} + \; a_{k + 1}.

So the recipe for Part 3 is:

Write the LHS for $n = k + 1$ and split off the $(k + 1)$ -th term.
Replace the remaining $k$ -term sum by the closed form from the hypothesis.
Factor out the common factor (almost always $(k + 1)$ ) and simplify until you reach the target formula with $n = k + 1$ in place of $n$ .

If the algebra gets messy, write down the target RHS at $n = k + 1$ first and steer your working towards it; knowing where you are heading makes the factoring obvious.

How exam questions ask about series identities

"Prove by mathematical induction that $\sum \dots = \dots$ ": the direct instruction. Reproduce all four parts; the verb "induction" tells you no other method earns the marks.
"Prove that $S_n = \dots$ for all positive integers $n$ ": the same task even though "induction" is not named, when $S_n$ is a sum. Use the four-part structure.
"Hence find $\sum_{i = 1}^{50} \dots$ ": a follow-up after you have proved the identity; substitute $n = 50$ into the closed form you just proved.
"Show that the result holds for $n = 1, 2, 3$ , then prove it in general": the first part is the base case (plus two extra checks for confidence); the "in general" part is the full induction.
"Prove $\sum_{i = 1}^{n} \frac{1}{i(i + 1)} = \frac{n}{n + 1}$ ": a telescoping or partial-fraction sum. Induction works exactly as for polynomial sums; split off the last term and simplify.

Exam technique

Lay the proof out under four clear labels (Base case, Assume $n = k$ , Prove $n = k + 1$ , Conclusion) so the marker can award each mark on sight. In the base case, evaluate the two sides separately and then state they are equal, rather than writing a single line that assumes they match. In the step, the line that earns the mark is the one where you substitute the hypothesis, so write "by the inductive hypothesis" explicitly at that point. When you factor out $(k + 1)$ , keep going until the expression is in exactly the same form as the original formula with $n$ replaced by $k + 1$ , and say so. Finish with the full conclusion sentence naming the principle of mathematical induction; a proof without it is marked as incomplete even when the algebra is perfect.

Worked example

Prove the sum of squares formula

Prove $\sum_{i = 1}^{n} i^2 = \frac{n(n + 1)(2 n + 1)}{6}$ for all positive integers $n$ .

Base case ( $n = 1$ ): The LHS is $1^2 = 1$ . The RHS is $\frac{1 \cdot 2 \cdot 3}{6} = 1$ . The two sides are equal, so the statement holds for $n = 1$ .

Inductive hypothesis: Assume the formula holds for $n = k$ , that is

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

Inductive step: Split off the last term and apply the hypothesis:

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

Factor out $(k + 1)$ and put everything over $6$ :

= (k + 1) \left[ \frac{k(2 k + 1)}{6} + (k + 1) \right] = (k + 1) \cdot \frac{k(2 k + 1) + 6(k + 1)}{6} = (k + 1) \cdot \frac{2 k^2 + 7 k + 6}{6}.

Factor the quadratic, $2 k^2 + 7 k + 6 = (k + 2)(2 k + 3)$ , so

\sum_{i = 1}^{k + 1} i^2 = \frac{(k + 1)(k + 2)(2 k + 3)}{6} = \frac{(k + 1)\big((k + 1) + 1\big)\big(2(k + 1) + 1\big)}{6}.

This is the formula with $n$ replaced by $k + 1$ , so the statement holds for $n = k + 1$ .

Conclusion: Since the base case holds and $P(k) \implies P(k + 1)$ , by the principle of mathematical induction the formula holds for all positive integers $n$ .

Prove a geometric series identity

Prove $1 + r + r^2 + \dots + r^{n - 1} = \frac{r^n - 1}{r - 1}$ for $r \neq 1$ and all positive integers $n$ .

Base case ( $n = 1$ ): The LHS is the single term $1$ ; the RHS is $\frac{r^1 - 1}{r - 1} = 1$ . Equal.
Inductive hypothesis: Assume $1 + r + \dots + r^{k - 1} = \frac{r^k - 1}{r - 1}$ .
Inductive step: Add the next term $r^k$ and use the hypothesis,

1 + r + \dots + r^{k - 1} + r^k = \frac{r^k - 1}{r - 1} + r^k = \frac{r^k - 1 + r^k(r - 1)}{r - 1} = \frac{r^{k + 1} - 1}{r - 1}.

This matches the formula at $n = k + 1$ . By the principle of mathematical induction the result holds for all positive integers $n$ .

Prove the sum of cubes formula

Prove $\sum_{i = 1}^{n} i^3 = \left( \frac{n(n + 1)}{2} \right)^2$ .

Base case ( $n = 1$ ): LHS $= 1$ , RHS $= \left( \frac{1 \cdot 2}{2} \right)^2 = 1$ . Equal.
Inductive hypothesis: $\sum_{i = 1}^{k} i^3 = \frac{k^2 (k + 1)^2}{4}$ .
Inductive step: Split off $(k + 1)^3$ and factor out $(k + 1)^2$ ,

\sum_{i = 1}^{k + 1} i^3 = \frac{k^2 (k + 1)^2}{4} + (k + 1)^3 = (k + 1)^2 \left[ \frac{k^2}{4} + (k + 1) \right] = (k + 1)^2 \cdot \frac{k^2 + 4 k + 4}{4} = \frac{(k + 1)^2 (k + 2)^2}{4}.

This is the formula at $n = k + 1$ , so by induction it holds for all positive integers $n$ .

Prove a partial-fraction (telescoping) sum

Prove $\sum_{i = 1}^{n} \frac{1}{i(i + 1)} = \frac{n}{n + 1}$ for all positive integers $n$ .

Base case ( $n = 1$ ): LHS $= \frac{1}{1 \cdot 2} = \frac{1}{2}$ , RHS $= \frac{1}{2}$ . Equal.
Inductive hypothesis: $\sum_{i = 1}^{k} \frac{1}{i(i + 1)} = \frac{k}{k + 1}$ .
Inductive step: Add the next term and combine over the common denominator $(k + 1)(k + 2)$ ,

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

This is $\frac{n}{n + 1}$ with $n = k + 1$ . By the principle of mathematical induction the identity holds for all positive integers $n$ .

Edge cases and what-ifs

A base case that is not $n = 1$ . If the identity is only claimed for $n \ge 2$ (for instance because a term is undefined at $n = 1$ ), verify the base case at $n = 2$ and run the step from $k \ge 2$ . The logic is unchanged.
Checking your closed form first. If you are unsure of the formula, evaluate the sum at $n = 1, 2, 3$ and confirm the closed form matches before you commit to a proof. The base case only checks one value, so a wrong formula can still pass it.
Telescoping as a cross-check. A sum like $\sum \frac{1}{i(i + 1)}$ also collapses directly because $\frac{1}{i(i + 1)} = \frac{1}{i} - \frac{1}{i + 1}$ . Induction is still the method the question asks for, but the telescoped value is a fast way to confirm your target RHS is right.
The empty sum. A sum from $i = 1$ to $i = 0$ is empty and equals $0$ . This is why a base case at $n = 0$ for $\sum_{i = 1}^{n} i = \frac{n(n + 1)}{2}$ also works: both sides are $0$ .

Common traps

Skipping the base case: Without the base, the chain has nothing to start, and the whole argument collapses. Always verify $P(1)$ explicitly, computing both sides separately.
Assuming what you are trying to prove: The hypothesis assumes $P(k)$ , not $P(k + 1)$ . Using $P(k + 1)$ inside the step is circular reasoning and earns no step marks.
Algebra errors in the step: Factoring out $(k + 1)$ then matching the target form is the reliable route. If the algebra is messy, write the target form for $n = k + 1$ first and aim at it.
Missing the conclusion sentence: The four-part structure includes a final "by the principle of mathematical induction" sentence. Markers expect it even when the algebra is complete.
Index confusion: $\sum_{i = 1}^{n}$ runs from $i = 1$ to $i = n$ . The $(k + 1)$ -term sum includes the term at $i = k + 1$ , which is exactly the new term you split off in the step.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

HSC-style5 marks

Prove by mathematical induction that for all positive integers

n

1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}

Show worked answer →

Base case ( $n = 1$ ): LHS $= 1$ , RHS $= \frac{1 \cdot 2}{2} = 1$ . Equal, so the statement holds for $n = 1$ .

Inductive hypothesis: Assume the statement holds for $n = k$ , that is

$1 + 2 + \dots + k = \frac{k(k + 1)}{2}$ .

Inductive step: Show the statement holds for $n = k + 1$ :

$1 + 2 + \dots + k + (k + 1) = \frac{k(k + 1)}{2} + (k + 1)$ by the hypothesis.

$= (k + 1) \left[ \frac{k}{2} + 1 \right] = (k + 1) \cdot \frac{k + 2}{2} = \frac{(k + 1)(k + 2)}{2}$ .

This is the formula with $n$ replaced by $k + 1$ . So the statement holds for $n = k + 1$ whenever it holds for $n = k$ .

Conclusion: By the principle of mathematical induction, the statement holds for all positive integers $n$ .

Markers reward each of the four parts (base, hypothesis, step, conclusion), the algebraic manipulation in the step, and the explicit citation of the inductive hypothesis.

Practise this

Sources & how we know this

Mathematics Extension 1 Stage 6 Syllabus (2017) — NESA
Mathematics Extension 1 HSC exam papers — NESA