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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

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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  ring?

Mar 18, 2015

#3
+95360
+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

Mar 19, 2015

#1
+17747
+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?

Mar 18, 2015
#2
+94558
+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   !!!!!

Mar 19, 2015
#3
+95360
+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