Proof by mathematical induction practice questions — IB Maths AA

${10}^3-10=990$, and $990\mathrm{/}6=165$. 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 $6$.

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

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.

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

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

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

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

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

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

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