finding a parimeter of a pattern ???

[THORTHELABRADOR]

The problem is at stage 0 a square's parimeter is 1 cm...stage 1 shows the origional square and a square on each side forming a cross...stage 2 is 5 rows as follows...1 sq,,3 sq, 5 sq, 3 sq 1 sq and stage 3 is as follows:
7 rows in this order....1 sq 3 sq, 5 sq, 7 sq, 5 sq,3 sq 1 sq....
i need to know the peramiter at stage 50 but how do I figure out the pattern to know what stage 50 looks like?
thanks
Thor the labrador

 

[tkhunny]

Find a pattern. Write out a description of each stage until you see it.

Stage 1 - One Square
Stage 2 - One on top, one on the bottom and one on each side for each existing row. 2 + 2 with now three existing rows. 1+2+2 = 5 total
Stage 3 - One on top, one on the bottom and one on each side for each existing row. 2 + 2*3 with now five existing rows. 1 + 2(2) + 2*1 + 2*3 = 13 total
Stage 4 - One on top, one on the bottom and one on each side for each existing row. 2 + 2*5 with now seven existing rows. 1 + 2(3) + 2*1 + 2*3 + 2*5 = 40 total

I'm beginning to see a pattern: For the n-th stage, maybe something like: 1 + 2(n-1) + 2(1 + 3 + 5 + ... + ???)

Can you decide where it stops? Stage 2 was 1, Stage 3 was 3, Stage 4 was 5. it seems to be increasing two each time. Hmmm....

Let's see what you get.

 

[THORTHELABRADOR]

I started to draw the pattern out and I see a difference of 4 at each stage. And I believe stage 50 should have 200 blocks in addition to the origional for a total of 201 blocks...I think...Let me try something

 

[tkhunny]

Four? You drew it incorrectly. Please have another go.

 

[BigGlenntheHeavy]

$\displaystyle Note: \ 0 \ is \ to \ 1, \ 1 \ is \ to \ 5, \ 2 \ is \ to \ 13, \ 3 \ is \ to \ 25, \ 4 \ is \ to \ 41, \ etc.$

$\displaystyle Hence, \ we \ have \ the \ sequence: \ 1,5,13,25,41,61,85,...$

$\displaystyle Using \ Newton's \ Finite \ Difference \ Formula, \ we \ arrive \ at \ f(n) \ = \ 2n^2+2n+1.$

$\displaystyle Ergo, \ f(50) \ = \ ?$

$\displaystyle For \ the \ perimeter, \ we \ have \ the \ sequence \ 1, \ 3, \ 5, \ 7, \ 9, \ ... \ cm.$

$\displaystyle Hence, \ f(n) \ = \ 1+2n, \ f(50) \ = \ 101 \ cm.$

 

[THORTHELABRADOR]

I redid it. I came up with a perimeter of 404.