Arithmetic Sequences & the Gauss Trick

1. 1Pair Sum Times Everything?A tempting shortcut: the first and last numbers make 21, so multiply 21 × 20 = 420. That answer is already sitting in the corner — do not trust it yet.21 × 20?
2. 2Even StepsWatch the +1 chip climb the steps: every step up adds exactly 1, no step more or less. The bars climb evenly from 1 to 20.step = +1
3. 3Fold: The First Pair FliesThe tallest bar (20) lifts off, flies over and lands stacked on the shortest (1). Two bars become one stack.1 + 20 = ?
4. 4One Pair, One SumThe number 21 grows out of the very bars that just folded — the sum of the first stack. So far: one pair.1 + 20 = 21
5. 5Fold Them All InFold on in the same way: the next outermost bar flies over onto its partner. Count with the counter in the corner — every stack that lands is one more pair.every pair = 21
6. 6Every Stack the Same HeightA guide line sweeps across the tops of all 10 stacks: every stack is exactly as tall as every other — because every stack is 21.10 stacks × 21
7. 7One BlockPush all 10 stacks together into one block: 10 stacks across, each 21 tall.10 × 21
8. 8The TotalTotal = pair sum × number of pairs = 21 × 10 = 210. The 21 grows out of a stack inside the block; the 10 comes from the pair counter.21 × 10 = 210
9. 9The Trap, Crossed OutNow deal with the trap answer 420 in the corner: to be true it would need 20 stacks — but the fold only made 10. The 10 missing stacks stand there as ghosts, so 420 is crossed out.21 × 20 = 420 ✗
10. 10Check by the Middle ValueAs the ghost stacks dissolve, check by a second road: the middle value of the staircase is 10.5 (half of 1 + 20) — the mid-line cuts every stack at exactly half its height. 10.5 × 20 numbers = 210. ✓ Same total.10.5 × 20 = 210 ✓
11. 11Before You FoldThe 10 × 21 block is the whole problem in one picture. Before folding any staircase, ask three questions: are the steps even? Is the number of terms even or odd? And what do the two outermost numbers make?(1 + 20) × 10 = 210

Change the numbers to test the engine with any sum & difference problem: