Solutions for Session 2, Part D

See solutions for Problems: D1 | D2 | D3 | D4 | D5

 Problem D1 One method is that the number of blocks for the nth staircase is n more than the number of blocks for the (n - 1)st staircase, because you need to add on a row of n blocks to build the new staircase. This allows you to construct a table. One clever trick is to piece together two consecutive staircases; the 5th staircase fits nicely into the 6th staircase to form a 6-by-6 square!

 Problem D2 There will be one new row in staircase n + 1, a row with n + 1 blocks in it. That means that to go from staircase n to staircase n + 1, you will need to add n + 1 blocks.

 Problem D3 In the 274th staircase there will be 37,401 + 274 = 37,675 blocks. In the 275th staircase there will be 37,675 + 275 = 37,950 blocks. This answer can also be reached by using (275) * (276) / 2 = 37,950.

 Problem D4 There are 5,050 blocks in the 100th staircase. The fastest way is to use n(n + 1)/2.

 Problem D5 A way to build a general formula is to notice that two copies of the 6th staircase will form a 6-by-7 rectangle. For the nth staircase this will be n(n + 1) blocks for two staircases, which means n(n + 1) / 2 for one.

 Session 2: Index | Notes | Solutions | Video