It's a good morning for some #ProblemSolving ! Here's today's Q from the November Calendar. What do you notice? What would your Ss notice? I'm going to #TryMathLive #TryMathsLive (thanks to @davidkbutler.bsky.social for this framing) #MTBoS #iTeachMath #MathSky #MathsToday
I'd be hoping to see some students drawing pictures -- perhaps if I made multilink cubes available -- leading to a nice hands-on proof.
I notice you have a bunch of difference of two squares :)
Isn't it lovely?
It is! We are starting a school-wide math challenge next week. We will post a problem each week and put it in our school newsletter (from old NCTM calendars). We will encourage students, staff, and families to submit solutions
Excellent!
I notice the base numbers increase from 1 to 199, each base is squared, and the signs alternate + and – I see a "difference of 2 squares" to start: 1^2 – 2^2 I know that can factor into (1+2)(1–2) --> 3•-1 The next pair 3^2 – 4^2 = (3+4)(3–4) --> 7•-1 Then 5^2 –6^2 --> 11•-1 Hmmm...
I can factor out those -1s. Leaves me with 3 + 7 + 11 + 15 would be next ... I can sort that out as a sequence but I'm lazy ... 😉 [I tell my students that mathematicians look for easy & elegant shortcuts] Remember that 3 was 1+2 and 7 was 3+4 and 11 was 5+6... So I have 1 + 2 + 3 + 4 + 5 + 6...
Ok that's a nice series to work with. How about the end? Before 199^2 must be 197^2 – 198^2 (197 + 198)(197 – 198) So my series of consecutive integers is 1 + 2 + ... 197 + 198; I need this sum times -1 plus 199^2 Good progress (so far without a calculator)
I remember that sum of an evenly spaced sequence (arithmetic series) is number of terms • average of first & last. So 198•(1 + 198)/2 198/2 • 199 rearrange for ease 99 • 199 [By the way, you can easily find a summation on TI84... it's in alpha-window shortcut menu or math menu]
Like I said, I'm doing this w/out calculator... my needed total is -1•99•199 + 199^2 Let's factor so I don't have to multiply... 199(-99 + 199) 199(100) 19900 Lovely! 😊 How did you work this out? For more problems, check out the calendar at karendcampe.wordpress.com/2024/11/01/n...
And the difference of two squares is one of my favorite noticings... it will factor into conjugates (a + b)(a – b). Conjugates pop up in so much later math (consider naming them for your Alg1 Ss) More on conjugates & "Powerful Pairs" in math: karendcampe.wordpress.com/2022/04/22/p...
Ok, stream of consciousness on what looks to be a great problem for high schoolers. It reminds me of the Basel problem, which I know converges, and it's alternating so it converges. Oh, silly, it's the integers not their reciprocals, and it's finite!
Ok, so it's all differences of squares, which will factor nicely. (1-2)(1+2)+(3-4)(3+4)..... The first factor will always be -1, so we can factor it out. (-1)*[ ( 1+2) + (3+4) + ...(198+199) ]. So we have just the negative of the sum of integers from 1 to 199.
Using Gauss' childhood trick, we have 100 (1+199) so a final result of -20000. I check this idea for 1-4+9-16 and it gives -10 which makes sense. Feeling self-satisfied but not 100% confident, I punch it into Desmos and get 19900. Ach! I've missed something!
I see it! My expression ends with a negative 199^2, but the series ends with a positive. I'll trim off the 199, find the sum, and then add it back. The sum should be (198/2)*(1+198)=19701 plus 199 gives me 19900, which I know should be -19900. A nice problem over Sunday breakfast!
Good work, but you retained a negative somewhere along the way... I believe the answer is +19900.
Yes! Funny, that. I found the correct answer, but poisoned it by pulling back in an earlier thought that it would be negative because you are always subtracting a larger number. I didn't reconsider this when I removed the 199^2.
Remove the 1^2 instead. Now it's all positive. Does that make it easier?
Sure. I thought of that later. 😀
I'm too lazy to reach for paper, but thinking difference of squares could come into play & could turn it into a geo or arithmetic series I can deal with (1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6) -3+-7+-11 Yup, arith with a1=-3, n=199, d=-4 (I forget the equation and am too left to figure it out/ look it up
Actually, that pattern ends at 198... Then add in 199²
(199²-198²)+...+(1²-0²) (199-198)(199+198)+...+(1-0)(1+0) 199+198+...+1+0 (199+0)+(198+1)+...+(100+99) 199×100 19900
Aaah you did it without those pesky negatives! Nice.