TASC 2024 TASC-style practice: Use mathematical induction to prove that 1 x 3 + 2 x 3^2 + 3 x 3^3 + ... + n x 3^n = (3/4)[(2n - 1)3^n + 1] for n >= 1.

Let P(n) be the statement that the sum equals (3/4)[(2n - 1)3^n + 1]. 1. Base case (n = 1). Left side = 1 x 3 = 3. Right side = (3/4)[(2(1) - 1)3^1 + 1] = (3/4)[3 + 1] = 3. So P(1) is true. 2. Inductive step. Assume P(k) is true: the sum to k terms = (3/4)[(2k - 1)3^k + 1]. 3. Add the (k + 1)th term, (k + 1)3^(k+1), to both sides. The new sum = (3/4)[(2k - 1)3^k + 1] + (k + 1)3^(k+1). 4. Write the 3^k term as (1/4)(2k - 1)3^(k+1) and combine over 3^(k+1): 3^(k+1)[(2k - 1)/4 + (k + 1)] = 3^(k+1)(6k + 3)/4 = (3/4)(2k + 1)3^(k+1). With the constant +3/4 this is (3/4)[(2(k+1) - 1)3^(k+1) + 1], which is P(k + 1). 5. Since P(1) holds and P(k) implies P(k + 1), by induction P(n) is true for all n >= 1. Markers reward a correct base case, an explicit inductive assumption, and clean algebra reaching the target form.

TASC 2022 TASC-style practice: Use mathematical induction to prove that 1/(1.2.3) + 1/(2.3.4) + ... + 1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1)(n+2)) for n >= 1.

Let P(n) be the given statement. 1. Base case (n = 1). Left side = 1/(1.2.3) = 1/6. Right side = (1 x 4)/(4 x 2 x 3) = 4/24 = 1/6. So P(1) is true. 2. Assume P(k): the sum to k terms = k(k + 3) / [4(k + 1)(k + 2)]. 3. Add the next term 1/[(k + 1)(k + 2)(k + 3)]. The new sum = k(k + 3)/[4(k + 1)(k + 2)] + 1/[(k + 1)(k + 2)(k + 3)]. 4. Use common denominator 4(k + 1)(k + 2)(k + 3): numerator = k(k + 3)^2 + 4 = k^3 + 6k^2 + 9k + 4 = (k + 1)^2(k + 4). 5. So the sum = (k + 1)(k + 4)/[4(k + 2)(k + 3)] = (k + 1)((k + 1) + 3)/[4((k + 1) + 1)((k + 1) + 2)], which is P(k + 1). 6. By the principle of induction P(n) holds for all n >= 1. The key step markers look for is factorising the cubic numerator to expose the (k + 1) factor.

TASSpecialist MathematicsSyllabus dot point

How can we prove a statement is true for every positive integer using induction?

Prove statements for all positive integers using the principle of mathematical induction.

The principle of mathematical induction, the base case and inductive step, and proofs for sums, divisibility and inequalities, with worked examples for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.79 min answerUpdated 2026-05-30

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

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What this dot point is asking

This dot point introduces the first formal proof technique of the course. Induction is the standard way to prove a statement that should hold for every positive integer, such as a formula for a sum, a divisibility fact, or an inequality. The logic is rigorous and the layout is expected to be precise, so examiners reward a clear, conventional structure.

The principle

The image is a line of dominoes: the base case tips the first one, and the inductive step guarantees that each falling domino tips the next, so they all fall.

Required layout

A complete induction proof should clearly state the proposition $P(n)$ , prove the base case, write down the inductive hypothesis explicitly, perform the inductive step ending at the $P(k+1)$ statement, and finish with a concluding sentence invoking the principle of induction. Missing any of these costs marks even when the algebra is correct.

Proving a summation formula

Worked example

Prove that $1 + 2 + 3 + \dots + n = \dfrac{n(n+1)}{2}$ for all integers $n \ge 1$ .

Let $P(n)$ be the statement that the sum of the first $n$ positive integers equals $\tfrac{n(n+1)}{2}$ .

Base case ( $n = 1$ ): the left side is $1$ and the right side is $\tfrac{1 \cdot 2}{2} = 1$ . So $P(1)$ is true.

Inductive step: assume $P(k)$ is true for some $k \ge 1$ , that is

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

Consider the sum up to $k + 1$ . Add $(k+1)$ to both sides:

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

This is exactly the formula with $n = k + 1$ , so $P(k+1)$ is true. By the principle of mathematical induction, $P(n)$ holds for all $n \ge 1$ .

Divisibility proofs

For divisibility, the trick in the inductive step is to write the $k+1$ expression in terms of the $k$ expression, so the inductive hypothesis can be substituted.

Proving a divisibility result

Worked example

Prove that $3^{2n} - 1$ is divisible by $8$ for all integers $n \ge 1$ .

Let $P(n)$ be the statement that $8 \mid (3^{2n} - 1)$ .

Base case ( $n = 1$ ): $3^2 - 1 = 8$ , which is divisible by $8$ . So $P(1)$ is true.

Inductive step: assume $3^{2k} - 1 = 8m$ for some integer $m$ . Consider $n = k + 1$ :

3^{2(k+1)} - 1 = 3^{2k}\cdot 9 - 1 = 9\cdot 3^{2k} - 1.

Write $9 \cdot 3^{2k} = 9(3^{2k} - 1) + 9$ , so

3^{2(k+1)} - 1 = 9(3^{2k} - 1) + 9 - 1 = 9(8m) + 8 = 8(9m + 1).

This is divisible by $8$ , so $P(k+1)$ is true. By induction, $8 \mid (3^{2n} - 1)$ for all $n \ge 1$ .

Inequality proofs

Inequalities use the same structure, but the inductive step adds or multiplies by a quantity rather than rearranging to an exact formula. State clearly each direction of inequality used.

Why this matters

Induction is the rigorous backbone for any result indexed by the positive integers, and the disciplined layout it demands carries over to all proof writing. Mastering the three proof types here, sums, divisibility and inequalities, covers the great majority of induction questions you will meet.

Exam-style practice questions

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

2024 TASC6 marksUse mathematical induction to prove that 1 x 3 + 2 x 3^2 + 3 x 3^3 + ... + n x 3^n = (3/4)[(2n - 1)3^n + 1] for n >= 1.

Show worked answer →

Let P(n) be the statement that the sum equals (3/4)[(2n - 1)3^n + 1].

Base case (n = 1). Left side = 1 x 3 = 3. Right side = (3/4)[(2(1) - 1)3^1 + 1] = (3/4)[3 + 1] = 3. So P(1) is true.
Inductive step. Assume P(k) is true: the sum to k terms = (3/4)[(2k - 1)3^k + 1].
Add the (k + 1)th term, (k + 1)3^(k+1), to both sides. The new sum = (3/4)[(2k - 1)3^k + 1] + (k + 1)3^(k+1).
Write the 3^k term as (1/4)(2k - 1)3^(k+1) and combine over 3^(k+1): 3^(k+1)[(2k - 1)/4 + (k + 1)] = 3^(k+1)(6k + 3)/4 = (3/4)(2k + 1)3^(k+1). With the constant +3/4 this is (3/4)[(2(k+1) - 1)3^(k+1) + 1], which is P(k + 1).
Since P(1) holds and P(k) implies P(k + 1), by induction P(n) is true for all n >= 1. Markers reward a correct base case, an explicit inductive assumption, and clean algebra reaching the target form.

2022 TASC7 marksUse mathematical induction to prove that 1/(1.2.3) + 1/(2.3.4) + ... + 1/(n(n+1)(n+2)) = (n(n+3))/(4(n+1)(n+2)) for n >= 1.

Show worked answer →

Let P(n) be the given statement.

Base case (n = 1). Left side = 1/(1.2.3) = 1/6. Right side = (1 x 4)/(4 x 2 x 3) = 4/24 = 1/6. So P(1) is true.
Assume P(k): the sum to k terms = k(k + 3) / [4(k + 1)(k + 2)].
Add the next term 1/[(k + 1)(k + 2)(k + 3)]. The new sum = k(k + 3)/[4(k + 1)(k + 2)] + 1/[(k + 1)(k + 2)(k + 3)].
Use common denominator 4(k + 1)(k + 2)(k + 3): numerator = k(k + 3)^2 + 4 = k^3 + 6k^2 + 9k + 4 = (k + 1)^2(k + 4).
So the sum = (k + 1)(k + 4)/[4(k + 2)(k + 3)] = (k + 1)((k + 1) + 3)/[4((k + 1) + 1)((k + 1) + 2)], which is P(k + 1).
By the principle of induction P(n) holds for all n >= 1. The key step markers look for is factorising the cubic numerator to expose the (k + 1) factor.

2023 TASC6 marksUse mathematical induction to prove that 1^2 x 2^1 + 2^2 x 2^2 + 3^2 x 2^3 + ... + n^2 x 2^n = 2^(n+1)(n^2 - 2n + 3) - 6 for n >= 1.

Show worked answer →

Let P(n) be the statement above.

Base case (n = 1). Left side = 1^2 x 2^1 = 2. Right side = 2^2(1 - 2 + 3) - 6 = 4 x 2 - 6 = 2. So P(1) is true.
Assume P(k): the sum = 2^(k+1)(k^2 - 2k + 3) - 6.
Add the (k + 1)th term (k + 1)^2 x 2^(k+1). New sum = 2^(k+1)(k^2 - 2k + 3) - 6 + (k + 1)^2 2^(k+1).
Factor 2^(k+1): 2^(k+1)[k^2 - 2k + 3 + (k + 1)^2] - 6 = 2^(k+1)[2k^2 + 4] - 6 = 2^(k+2)[k^2 + 2] - 6.
The target is 2^(k+2)((k+1)^2 - 2(k+1) + 3) - 6, and (k+1)^2 - 2(k+1) + 3 = k^2 + 2, matching exactly. So P(k + 1) holds.
By induction P(n) is true for all n >= 1. Markers reward the base case, a stated assumption, and the algebra that collapses k^2 - 2k + 3 + (k+1)^2 to 2(k^2 + 2).