 Quandaries and Queries if i make a block pyramid and it puts a new perimeter around it every time, for example the first layer will be 1 block across (area=1), the second layer will be 3 blocks across (area=9), the third layer will be 5 blocks across (area=15),etc. The normal block pyramid. I have figured out that in order to figure out the number of blocks needed for a certain level, the equation is (2x-1)2 or (2x-1)(2x-1), where x is equal to the level. For example, on the fourth level, the equation tells you that it will have an area of 49. How would i make an equation for the total number of blocks up to the level. For example, in order to complete 1 level you need 1 block, for 2 levels you need 10 blocks, for three levels you need 35 blocks, and for 4 levels you need 84 blocks. Thank you for considering this. Sincerely, Kyle Hi Kyle, To find the total number of blocks needed to build your pyramid to the level x you need to sum the squares of the odd integers from 1 to 2x - 1. The key to solving this problem is to know how to find the sum of all the integers from 1 to n. This expression is 12 + 22 + 32 + 42 +···+ n2 = n(n+1)(2n+1)/6 I am going to write the sum on the left with n = 2x and then reorder it to get the sum of the squares of the odd integers plus the sum of the squares of the even integers, that is 12 + 22 + 32 + 42 +···+ (2x)2 = [12 + 32 + 52 +···+ (2x - 1)2 ] + [22 + 42 + 62 + (2x)2 ] so 12 + 32 + 52 +···+ (2x - 1)2 = [12 + 22 + 32 + 42 +···+ (2x)2 ] - [22 + 42 + 62 + (2x)2 ] = [12 + 22 + 32 + 42 +···+ (2x)2 ] - 4 [12 + 22 + 32 +···+ (x)2 ] Now apply the expression for the sum of all the integers from 1 to n to get 12 + 32 + 52 +···+ (2x - 1)2 = [ (2x)(2x+1)(4x+1)/6 ] - 4 [ x(x+1)(2x+1)/6 ] I simplified this expression to get 12 + 32 + 52 +···+ (2x - 1)2 = x(2x-1)(2x+1)/3 Penny   