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Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains 8 unit squares. The second ring contains 16 unit squares. If we continue this process, then what is the number of unit squares in the $100^{th}$ ring?

 fill((2,2)--(7,2)--(7,7)--(2,7)--cycle,gray(0.8)); fill((3,3)--(6,3)--(6,6)--(3,6)--cycle,gray(0.6)); fill((4,4)--(4,5)--(5,5)--(5,4)--cycle,black); for (int i=0; i<10; ++i) { draw((0,i)--(9,i)); draw((i,0)--(i,9)); }

Guest Mar 18, 2015

Best Answer 

 #3
avatar+92806 
+10

I saw a different pattern again

8,16,20,.....

I saw 16 as  3*4+4     (I added the corner blocks seperately)

I saw 20 as   5*4+4    (again I added the corner blocks afterwards.)

Looking at the patter I would get  8=1*4+4   that's true, great,

so I have

 

n                                 1           2            3

number of squares    1*4+4,  3*4+4,   5*4+4,  ...... 

now lets look at a pattern for 1,3,5 etc ...    they are getting bigger by 2s ....    mmm,    (2n-1)

So   

$$\\T_n=(2n-1)*4+4\\
T_n=4(2n-1)+4\\
T_n=4[(2n-1)+1]\\
T_n=4[2n]\\
T_n=8n\\$$

so

$$T_{100}=8*100=800 squares$$

 

Yet another proof that

ALL ROADS LEAD TO ROME.        (๑‵●‿●‵๑)  

Melody  Mar 19, 2015
 #1
avatar+17744 
+10

The first ring has 8 squares           = 1 x 8

The second ring has 16 squares   = 2 x 8

The third ring has 24 squares       = 3 x 8

Continuing with this pattern, can you find the number of squares in the 100th ring?

geno3141  Mar 18, 2015
 #2
avatar+87333 
+10

Very nice, geno....

I happened to notice another pattern, too

1st ring = (3^2 - 1^2)  = 8

2nd ring (5^2 - 3^2)  = 16

3rd ring (7^2 - 5^2) = 24

4th ring (9^2 - 7^2)  = 32

So the nth ring is just [(2n + 1)^2 - (2n -1)^2 ]  =   8n 

The same result as geno's   !!!!!

 

  

CPhill  Mar 19, 2015
 #3
avatar+92806 
+10
Best Answer

I saw a different pattern again

8,16,20,.....

I saw 16 as  3*4+4     (I added the corner blocks seperately)

I saw 20 as   5*4+4    (again I added the corner blocks afterwards.)

Looking at the patter I would get  8=1*4+4   that's true, great,

so I have

 

n                                 1           2            3

number of squares    1*4+4,  3*4+4,   5*4+4,  ...... 

now lets look at a pattern for 1,3,5 etc ...    they are getting bigger by 2s ....    mmm,    (2n-1)

So   

$$\\T_n=(2n-1)*4+4\\
T_n=4(2n-1)+4\\
T_n=4[(2n-1)+1]\\
T_n=4[2n]\\
T_n=8n\\$$

so

$$T_{100}=8*100=800 squares$$

 

Yet another proof that

ALL ROADS LEAD TO ROME.        (๑‵●‿●‵๑)  

Melody  Mar 19, 2015

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