it can be solved wid sequence nd series as well…
@levi: You got it! Tanks for sharing.
Hi Kaled, Based on this same analogy I found a a-ha moment to describe (a+b)^2 = a^2 + b^2 + 2ab. Keep a square and b square next to each other. and we would need two rectangles (one horizontal and one vertical) with sides a and b to complete a+b square… Thanks to you
@Melissa: Thanks, I love it when counting can become fun :). I find it can help to visualize math with real-world objects.
As far as the cubes go (as well as higher dimensions) take the series 1,8,27,64,125…
take the differences (7,19,37,61) take the differences again (12,18,24…) take the differences again (6,6…) and notice that you have 3! (just as you had 2! with the squares).
raise to the fourth power, and you will have to take the differences of one more generation before you end up with 24 or 4!
The pattern continues thus…
I discovered this when I was bored in shul one shabbos, and I had to sneak into a backroom to use a pen
In the case of the cube, you can see growth the of x^3 geometrically as extending by x^2 on three sides then 3x fills all the rest of the space except for one corner. This matches up with the algebraic expansion of (x+1)^3 = x^3 + 3x^2 + 3x + 1.
Its a bit harder to visualize in four dimensions, but you get the point
I never thought about using it with pebbles either it makes things so much clearer! Really interesting
wow, this is great ! I have never heard of some of these methods and all the different things you can do with calculus concerning patterns
Good way to look at squared numbers with the pebbles. I never knew there was so much to sqaured numbers
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write out all squares of 0-50 and you’ll notice the last two digits make a pattern
Now counting down from 50 squared
recognize the pattern?
this is the way that i made up
2 2 2 2
9 +(3+(9-1)x2)=10 or 10 + (3+(10-1)x2)=11 do you see the pattern
Apologies if i’ve committed some heinous error below - it’s been some time since I had to use algebra but…
given the obvious analogs between the cubes and the squares, could we generalise the result for an X-dimensional object?
This works out for a 2 dimensional object as
0 = Xn^0 = 1
1 = Xn^1 = 2n
This works out for a 3 dimensional object as
0 = Xn^0 = 1
1 = Xn^1 = 3n
2 = X*n^2 = 3n^2
(hence 3n^2 + 3n + 1)
and for a 4 dimensional object we might then project this would be
0 = Xn^0 = 1
1 = Xn^1 = 4n
2 = Xn^2 = 4n^2
3 = Xn^3 = 4n^3
(hence 4n^3 + 4n^2 + 4n + 1)
I found this out at school and i found another pattern
if we add odd continously (consicutivly) the no. of odd numbers equal to the square root of the sum of the number
example - 1+3+5+7+9+11+13+15+17+19=100
squre root of 100 is equal to 10
= 1,3,5,7,9,11,13,15,17,19 are 10 numbers
awesome awesome awesome i just loved your square and square root patterns
and had a great help in my holidays homework…
THANKS A LOT!!!
I really enjoyed this article; well done!
Something my math teacher showed me:
In other words:
1st odd number=1^3
sum of next 2 odd numbers=2^3
sum of next 3 odd numbers=3^3
sum of next 4 odd numbers=4^3
…and so on.
I haven’t looked at this long enough to be able to say “I can’t figure out why it works,” but for now all I see is that:
the average of all of the numbers being summed = (# of numbers being summed)^2, and hence,
(the average of all of the numbers being summed)^(3/2)=the sum of the numbers