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Freeman

Equation with squares???

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OK, I have this problem stuck in my mind and I need to know the equation. There is a square consisting of six rows of six squares, thus thirty six miniature squares inside that one, but how many squares can be formed from that big square??? Does anyone know the formula??? :confused:

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[math]\sum_{X=0}^{n-1}{(X+1)^2}[/math]

 

That's seems about right.

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My memory is rusty, how would I solve this? I plug in for x, and add all results togather?

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think it comes out to be n(n+1)(2n+1)/6 . i ve asked this question before in another thread

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Yeah, I didn't do that quite right. I think after a quick edit it'll be right.

 

Take the number of squares in each row and call that n. Then, starting with with X=0, do (x+1)^2. After that, add 1 to X and repeat the process. Continue doing so until X=n.

 

So for a square with 2 in a row:

 

(0+1)2=1

(1+1)2=4

Now, X=1.

This is one less than n (as asked for by the n-1 part). Therefore you stop and add all the terms together for a total of 5.

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I don't know if I did this correctly, lemme post it here...

[math](0+1)^2=1[/math]

[math](1+1)^2=2^2=4[/math]

[math](2+1)^2=3^2=9[/math]

[math](3+1)^2=4^2=16[/math]

[math](4+1)^2=5^2=25[/math]

[math](5+1)^2=6^2=36[/math]

[math]36+25+16+9+4+1=10+20+25+36=55+36=91[/math] So 91 possibilities? Correct?

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I think i may have a typo in the answer in the other thread. instead of summing k^2 i summed k

 

take a rby r square where r<=n

 

now if we look at the possible bottom left corner of the square on the grid, we can see that there are (n-r+1)2 positions the bottom left corner can take place.

 

so for square size r=n the total no of square = 1

for r=1 total number is = n^2

 

so bassically the total number is

 

sum (n-r+1)2 from r = 1 to r = n . ie all the possible sizes of squares. thats equivalent to sum of 1^2 +...+n^2=n(n+1)(2n+1)/6 squares in total.

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So, wait, the answer isn't 91 ?!?

Wait, if r=n and r=1, wouldn't n=1???

so, what???

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i dont understand you

 

if u want to find the number of squares of a certain size in the grid of n by n unit squares i.e size r, then the number is (n-r+1)^2. so for r = 1 we get n^2 squares and for r=n we get only 1 square which is obvious. to find the total number of squares of all sizes we just sum up the (n-r+1)^2 bit for all the possible sizes of squares. i.e from r=1 to r=n

which is the same as n^2 + (n-1)^2.......+1^2

 

equivalent to sum of k^2 from k=1 to k=n, which is equal to n(n+1)(2n+1)/6

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91 is the right answer. What I'm not sure about is what "r" is. Bloodhound said up there something about "take a rby r square where r<=n" and I didn't follow that. Also, I don't understand how you got from "n^2 + (n-1)^2.......+1^2" to "which is equal to n(n+1)(2n+1)/6" that.

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here we have a grid made out of n by n "unit" squares. if u want to find the number of r by r square, we obviously must have r<=n , cos there are no squares in the grid bigger that n by n.

 

now if we take a 1 by 1 square. and consider its bootom left corner. we try to find all the possible places to put the square in the grid. the total number of places is n^2

 

for a 2 by 2 square. the total number of places is (n-1)^2

 

and for a r by r square. the total number of places is (n-r+1)^2

 

so to find the total number of squares in the grid of all possible sizes.

 

we sum (n-r+1)^2 for all possible values of r.

 

i.e we find the number of square of 1 by 1 and add it to number to squares of 2 by 2 and add all the way to the number of squares of n by b.

 

this sum comes out to be

[math]\sum_{r=1}^{n}{(n-r+1)^2}[/math]

which when u work out. comes to be

[math]\sum_{k=1}^{n}{k^2}[/math]

 

this is a standard series, whos nth partial sum can be expressed explicitely.

 

and its equal to n(n+1)(2n+1)/6

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