Proof by mathematical induction — IB Maths AA practice questions

Question 1

By induction one can show that $n^3 - n$ is divisible by $6$ for every $n \in \mathbb{N}$. Compute $\dfrac{10^3 - 10}{6}$.

Accepted Answer

$10^3 - 10 = 990$, and $990 / 6 = 165$. The induction proof factors $n^3 - n = (n-1)n(n+1)$, which is the product of three consecutive integers and is therefore divisible by both $2$ and $3$, hence by $6$.

Answer

$165$

Answer

$166$

Answer

$100$

Answer

$200$

Question 2

Show by induction that $1 \cdot 2 + 2 \cdot 3 + \cdots + n(n+1) = \dfrac{n(n+1)(n+2)}{3}$ for every $n \ge 1$. Using this, compute $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5$.

Accepted Answer

$\dfrac{4 \cdot 5 \cdot 6}{3} = \dfrac{120}{3} = 40$. (Direct: $2 + 6 + 12 + 20 = 40$.)

Answer

$30$

Answer

$40$

Answer

$50$

Answer

$60$

Question 3

Which statement is logically equivalent to "if $n^2$ is even, then $n$ is even"?

Accepted Answer

The contrapositive of "$n^2$ even $\Rightarrow$ $n$ even" is "$n$ odd $\Rightarrow$ $n^2$ odd". A statement and its contrapositive always have the same truth value.

Answer

If $n$ is even, then $n^2$ is even

Answer

If $n$ is odd, then $n^2$ is odd

Answer

If $n^2$ is odd, then $n$ is even

Answer

$n$ is even if and only if $n^2$ is odd

Question 4

By induction, $1 + 3 + 5 + \cdots + (2n - 1) = n^2$. Compute this sum for $n = 9$.

Accepted Answer

$1 + 3 + 5 + \cdots + 17 = 9^2 = 81$. (Direct check: $1+3+5+7+9+11+13+15+17 = 81$.)

Answer

$49$

Answer

$64$

Answer

$81$

Answer

$100$

Question 5

The negation of "for every real number $x$, $x^2 + 1 > 0$" is:

Accepted Answer

Negation of $\forall$ is $\exists$; negation of $>$ is $\le$. So the negation is $\exists x \in \mathbb{R}:\ x^2 + 1 \le 0$. (This statement is itself false, but the question asked for the negation, not its truth value.)

Answer

For every real $x$, $x^2 + 1 \le 0$

Answer

There exists a real $x$ such that $x^2 + 1 \le 0$

Answer

There exists a real $x$ such that $x^2 + 1 > 0$

Answer

For every real $x$, $x^2 + 1 < 0$

Question 6

Assume $1 + 2 + \cdots + k = \dfrac{k(k+1)}{2}$ (induction hypothesis). To complete the inductive step we must show that $1 + 2 + \cdots + (k+1)$ equals:

Accepted Answer

The target formula at $n = k+1$ is $\dfrac{(k+1)\big((k+1)+1\big)}{2} = \dfrac{(k+1)(k+2)}{2}$. (And indeed $\dfrac{k(k+1)}{2} + (k+1) = \dfrac{(k+1)(k+2)}{2}$, completing the step.)

Answer

$\dfrac{(k+1)(k+2)}{2}$

Answer

$\dfrac{k(k+1)}{2} + 1$

Answer

$\dfrac{(k+1)^2}{2}$

Answer

$(k+1)(k+2)$

Question 7

The formula $1^3 + 2^3 + \cdots + n^3 = \left[\dfrac{n(n+1)}{2}\right]^2$ holds for all $n \ge 1$. Compute $1^3 + 2^3 + 3^3 + 4^3$.

Accepted Answer

$\left[\dfrac{4 \cdot 5}{2}\right]^2 = 10^2 = 100$. (Verify directly: $1 + 8 + 27 + 64 = 100$.)

Answer

$100$

Answer

$144$

Answer

$36$

Answer

$1024$

Question 8

Let $p$: "every prime number is odd" and $q$: "$3 + 4 = 7$". The truth value of $p \wedge q$ is:

Accepted Answer

$p$ is **false** (since $2$ is prime and even), $q$ is **true**. From the truth table for conjunction, $	ext{false} \wedge 	ext{true} = 	ext{false}$.

Answer

True

Answer

False

Answer

Cannot be determined

Answer

Both true and false

Question 9

Verify the base case $n = 1$ for the formula $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2}$. What is the value of the right-hand side at $n = 1$?

Accepted Answer

At $n = 1$: $\dfrac{1 \cdot 2}{2} = 1$, which equals the left-hand side $1$. The base case holds.

Answer

$0$

Answer

$1$

Answer

$2$

Answer

$n + 1$

Question 10

Let $A = \{1, 2, 3, 5\}$ and $B = \{2, 3, 5, 7\}$. Then $A \cap B$ equals:

Accepted Answer

$A \cap B$ contains the elements common to both sets: $\{2, 3, 5\}$.

Answer

$\{1, 2, 3, 5, 7\}$

Answer

$\{2, 3, 5\}$

Answer

$\{1, 7\}$

Answer

$\emptyset$

Proof by mathematical induction practice questions — IB Maths AA

Problems & worked solutions

Problem 1 — Proof by mathematical induction

Problem 2 — Proof by mathematical induction

Problem 3 — Proof by mathematical induction

Problem 4 — Proof by mathematical induction

Problem 5 — Proof by mathematical induction

Problem 6 — Proof by mathematical induction

Problem 7 — Proof by mathematical induction

Problem 8 — Proof by mathematical induction

Problem 9 — Proof by mathematical induction

Problem 10 — Proof by mathematical induction