As my first contribution, I hope this one will be curious enough for you.
Do you remember about one of my inquiries in class regarding the easiest thing that anyone can do in Maths?
It was, of course, TO COUNT. Therefore, if you could reduce in some way another kind of problem to a counting-problem, it would be easier. The purpose of this entry is to put an example about this way of reasoning:
"What is the sum of all the numbers from 1 to 500?"
Whenever I mentioned this arithmetic problem in class, it seemed to take too long due to the quantity of sums it required. So let's have a look at another way to figure out the result of the sum:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + ... (here go all natural numbers between) ... + 497 + 498 + 499 + 500.
If you were counting some things (red dots, for example) that have to do with that situation, they could be displayed as groups by rows, with each group standing for each one of these numbers, and all groups aligned at their beginning, like this:
• (1)
• • (2)
• • • (3)
• • • • (4)
• • • • • (5)
(etc.)
It is very easy to realize that all groups form a right triangle -with same height as length in dots, so it is isosceles too-, no matter how many numbers you want to consider, in our case 500. And that is very near to be half square with 500 dots on each side.
Now, imagine every dot inside a square box, and add more boxes on each row to complete a big square. It is obvious that we have 500 · 500 boxes, the only thing left to decide is how many of them have a red dot. It will help if we colour diagonal boxes in blue, and other boxes with dots in yellow, like in the picture below.
Boxes in both colours match to those with a dot. Notice that blue boxes must be exactly 500.
And what about yellow ones?. Notice that there are same number of yellow boxes as no-color boxes, so they are
(500 · 500 - 500) : 2 = 124 750 .
Hence, the red dots are 125 250.
By the way... ...Could you apply this to other examples?
This comment has been removed by the author.
ReplyDeleteI don´t understand this , Rafa.
ReplyDeleteI think that we must see this in class for understand better.
Bye :)
Hello!
ReplyDeleteThe sum of all the numbers (from 1 to 250)is:
(250x250):2=31.250
I think it's correct.
Bye!
I was thinking and these isn't the solution.
ReplyDeleteI have made an easy example:
1+2+3+4+5=15 (5x5+5):2=15
So:(250x250+250):2=31375
Hi!:)
ReplyDeleteI don't understand the question...
Bye! :)
Hello!
ReplyDeleteThe sum of all the numbers from (1 to 110)
(110·110-110):2= (12100-110):2= 11990:2=5995
Bye.
Hello class!
ReplyDeleteThe sum of all the numbers from 1 to 50.
(50·50-50):2 = (2500-50):2 = 2450:2 = 1225
The sum of all the numbers from 1 to 50 is 1225.
Bye class!
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteYou got it (or almost)!
ReplyDeleteRafa Martinez and I were discussing about it was a too complex question for 2º ESO students. But some of you are very near, and even correct, as I see in examples you mentioned above.
My trick consists on subtracting from the squared highest number in the sequence, that proper number, then obtaining the half of that, and after this adding the number again...
But you inferred a way to simplify that a bit more!
Regarding your examples:
Sum from 1 to 5 is [my way]
(5·5 - 5): 2 + 5 =
(25 - 5):2 + 5 =
20 : 2 + 5 = 10 + 5 = 15
[Your way] (5·5 + 5): 2 = 15 too!
Can anyone explain why are both ways driving to same result, no matter the number considered?
Sum from 1 to 250 is essentially correct. Congrats J.A.!
Rafa: The question is to sum consecutive numbers from one (1) to any you want to, in the simplest way you can find!
Ana Maria and Javi: Try again with cases you selected! You're very close to! There's only a detail missing!
hello!
ReplyDeleteSum of 1 to 20
(20.20-20):2+20=
(400-20):2+20=
380:2+20=
190+20=210
and...
1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20=210
This is ok!
Bye!
1to 20
ReplyDeleteHello!
ReplyDelete1 to 30=
=(30·30-30):2+30=
=(900-30):2+30=
=870:2+30=
=435+30=
=465
Goodbye:)
Hello!
ReplyDelete1 to 110=
(110·110-110):2+110= (12100-110):2+110= 11990:2+110= 5995+110= 6105
I think it is correct!
Hello!
ReplyDelete1 to 45=
(45·45-45):2+45=
=(2025-45):2+45=
=1980:2+45=
=990+45=
=1035
Ok! So I'll try to do now. What are the sum for all the numbers 1 to 150?
ReplyDelete(150·150+150)/2 = (22500+150)/2 = 22650/2 = 11325
I don't understand the question, if I remember I'm going to tell you in the class okay¿?
ReplyDeleteI think it's corret
ReplyDelete1 to 110=
(110·110-110):2+110= (12100-110):2+110= 11990:2+110= 5995+110= 6105
bye class