Perimeter Of Cross - Way Out There

The equation to this problem is 8n-4=perimeter of the figure n being the figure number. Therefore 8x20-4=156

I analyzed the crosses from fig. 1-4 and I realized that their perimeters have pattern. In 1, it's 4x1. In 2, it's 4x3. In 3, it's 4x5. In 4, it's 4x7 and so on until 20. Following the pattern I made the formula 2k+1 that k as the lower limit having the value of zero. and my higher limit is 19 which is for fig. 20. 2(19)+1 = 39. Finally, we have to multiply 39 by 4 getting 156 :D

It forms an A.P. with first term=a, n= 20 and difference=8

a+(n-1)d=4+(20-1)8=156

To find the number of boxes: (where n is the number of boxes and x is the number it is in the series , e.g. fig.1 is 1) n=4(x-1)+1 : this gives 77 boxes for x=20 : then apply general formula : where n is the number of boxes and P is the perimeter : P =2n+2 : giving a final answer of 156

It is simple.Actually there is a series osf increasing 8.So the answer =8n-4=8*20-4=156.

after fig 2 theres increase in 3 units per side and loss of 1 unit per side as the fig goes on at fig 20 we are left with 39 units each side then 39*4 that will give u 156