I came across this fun pattern, can anyone give me a proof for this?

Question

The Pattern

$\begin{align*}1(8)&=(3^2)-1\\ 2(8)&= (3+1)^2 \\ 3(8)&= (3+2)^2-1\\ 4(8)&= (3+3)^2 - 4\\ 5(8)&= (3+4)^2- 9\\ 6(8)&= (3+5)^2-16\end{align*}$

Conjecture

I think the pattern is that the numbers appear in the following form:

$$8(n+1) = (3+n)^2 - (n-1)^2$$

Please correct me if I'm wrong, I'm not really that good at math.

@Peter yes. couldnt see that because my notebook was filled with scribbles — Satvik Shubh, Jan 18 '23 at 11:55
Plug $n$ in the previous formula I mention you will find the reason why? It's actually an identity — , Jan 18 '23 at 11:56
Just expand the squares using the binomial formula and simplify — algebroo, Jan 18 '23 at 12:02
For what its worth, this is not a result of number theory but rather is a result of highschool algebra. — JMoravitz, Jan 18 '23 at 13:25
Congratulations on your discovery. That said, before you ask a question here you should rewrite the scribbles in your notebook on a new clean page. Then think about how to understand the pattern you observed. See if anything you already know (like ordinary algebra) helps. — Ethan Bolker, Jan 18 '23 at 16:38
@SatvikShubh: Your (fixed) identity is true. Congratulations on observing and describing a pattern! That's what mathematics is all about. Keep it up! — Blue, Jan 18 '23 at 16:40
@EthanBolker Thank you. I find disorganized notes more helpful (idrk why) but yeah ill present it to my teachers/math department in a neater way — Satvik Shubh, Jan 18 '23 at 16:59
@SatvikShubh You always start disorganized. Then you learn more when you clean it up. — Ethan Bolker, Jan 18 '23 at 18:29

score 2 · Answer 1 · answered Jan 18 '23 at 16:46

If you are just learning algebra, this might not be obvious. But, the rule to keep in mind is that multiplication distributes over addition. On the left-hand side, you have $8(n+1)$ which after distribution equals $8n+8.$ Now we have to show that the right-hand side gives the same thing.

Let's work out the two terms one at a time.

$(n+3)^2 = (n+3)(n+3) = n(n+3) + 3(n+3) = n^2 + 3n + 3n+9 = n^2+6n+9$

$(n-1)^2 = (n-1)(n-1) = n(n-1) + (-1)(n-3) = n^2 - n - n +1 = n^2-2n+1$

$(n+3)^2 - (n-1)^2 = (n^2+3n+9)-(n^2-2n+1) = n^2+6n+9 - n^2+2n-1 = 8n+8$

As required.

Hope this helps

Yup. asked my teacher and she said the same thing. Thanks! – Satvik Shubh Jan 18 '23 at 16:57 — Satvik Shubh, Jan 18 '23 at 16:57

I came across this fun pattern, can anyone give me a proof for this?

1 Answers1