ALGEBRAIC OPERATIONS

Each time, a number is multiplied by the number 2 more than itself.

$$n(n+2)$$

$10(10+2)=10\times12=120$. Direct check: $10\times12=120$ ?.

$$2n+(2n+2) = 4n+2$$

$n=1$: sum $=2+4=6$. Is 6 a multiple of 4? No. So the original claim in the question title is FALSE — $4n+2$ is actually always 2 more than a multiple of 4, not a multiple of 4 itself.

$$n+(n+1)+(n+2)+(n+3)+(n+4) = 5n+10$$

$$5n+10 = 5(n+2)$$

$6+7+8+9+10=40=5\times8$ ?, confirming the claim (and matching $5(n+2)$ with $n=6$: $5(8)=40$).

$n=0$: $0+0+41=41$ (prime). $n=1$: $1+1+41=43$ (prime). $n=2$: $4+2+41=47$ (prime).

Testing specific numbers can only show a pattern HOLDS for those cases — it can never prove the pattern holds for ALL possible values, since there are infinitely many numbers to check. A single counterexample (and one does exist for this formula, at $n=40$, where $n^2+n+41=41^2$, which is NOT prime) is enough to disprove a general claim, no matter how many earlier cases worked.

Using either data point: $k=1.5/3=0.5$ (checking: $0.5\times7=3.5$ ?). So $$d=0.5t$$

$d=0.5(60)=30$ m. This seems reasonable for a factory conveyor belt moving at a slow, steady pace over a full minute.

$$(2n+1)^2+(2n+3)^2 = 4n^2+4n+1+4n^2+12n+9 = 8n^2+16n+10$$

$$8n^2+16n+10 = 8(n^2+2n+1)+2 = 8(n+1)^2+2$$

$5^2+7^2=25+49=74=8(9)+2$ ?.

$n=2$: $2$. $n=3$: $6$. $n=4$: $12$. $n=5$: $20$. $n=6$: $30$.

$n^2-n=n(n-1)$ — this is a product of TWO CONSECUTIVE integers. Since any two consecutive integers must include exactly one even number, their product must always be even (an even number times anything is always even).

$4\times5\times6=120=6\times20$ ?. $7\times8\times9=504=6\times84$ ?.

Among any 3 consecutive integers, AT LEAST one must be even (divisibility by 2 guaranteed), and exactly one out of every 3 consecutive integers must be a multiple of 3 (since multiples of 3 occur every third number, guaranteeing divisibility by 3). Since the product is guaranteed divisible by both 2 AND 3 (which share no common factor), it must be divisible by their product, $2\times3=6$.

$n=1$:3, $n=2$:7, $n=3$:13, $n=4$:21, $n=5$:31 — all calculated correctly. However, $21=3\times7$ is NOT prime — the student's claim that 'all of these look prime' was actually incorrect even for the tested values, showing they didn't check carefully enough.

Testing only a handful of small cases can create a false sense of confidence — as shown here, even the student's OWN test cases contained a counterexample they missed. A pattern appearing to hold for several values is never a substitute for genuine proof; formulas that produce primes for many small inputs (like the famous $n^2+n+41$) eventually fail for larger $n$, which is why rigorous verification of every single claimed case matters, not just a quick glance.

$R(6)=2(36)+30+3=72+30+3=105$ thousand. $R(12)=2(144)+60+3=288+60+3=351$ thousand.

$R(60)=2(3600)+300+3=7200+300+3=7503$ thousand (over \$7.5 million). This is risky because a quadratic model grows increasingly steeply (the growth RATE itself keeps increasing) — real business revenue growth typically slows or plateaus over time due to market saturation, competition, or capacity limits, so a model fitted to early data is very unlikely to remain accurate when extrapolated many years into the future; the quadratic's ever-accelerating growth is not realistic indefinitely.

$$R(15)=18\times15=270\text{mm}$$

Solving $2500\le18n\le3200$: $n\ge138.9$ and $n\le177.8$, so $n$ between 139 and 177 (whole steps) satisfies the regulation. A 170-step staircase, while mathematically satisfying the formula and even falling within this range, would be wildly unrealistic in practice — real staircases within a single flight rarely exceed 20-30 steps; a 170-step single flight would be a very unusual (likely a stadium or monumental structure, not standard building regulation context), showing that a formula being mathematically valid doesn't guarantee every output value is practically sensible.