Mathematical reference
- Counting entries accumulate
- 1+2+3+4+5 = 15
- Triangular entries accumulate
- 1+3+6 = 10
- Six triangular layers
- 1+3+6+10+15+21 = 56
Lesson transcript
This is the text of the narrated lesson. References to turning, highlighting or changing an object describe the linked interactive lesson; its demonstrations are not embedded in this reading page.
Read down a fixed diagonal
Begin with a diagonal just inside the left edge of Pascal’s Triangle. Follow it downward: one, two, three, four, and onward. Here a diagonal keeps the same position within each successive row. This is different from reading horizontally across one row. The selected four lies in row four, position one.
Move one step down that same diagonal to five. Its position is still one, while its row has increased to five. The highlighted contributors along the outside edge are five ones. Adding those ones gives five. The new diagonal is collecting values from the preceding one, beginning at its first boundary entry.
Continue to six. Another outside one joins the collection, so the accumulated total rises from five to six. The local two-parent rule explains that step: the previous five and the new boundary one give six. A whole diagonal of accumulating totals can therefore be traced back to the parent rule.
Pull back with six still selected. You can read the outside diagonal as repeated ones, and the next diagonal as their running totals. We will repeat that operation farther inward. Keep track of the diagonal providing the contributions and the neighboring diagonal receiving the total; their roles are related but distinct.
Let the contributions accumulate
Return to five and clear the highlighted contribution list. The destination remains five: it is an existing entry in the triangle. What we have reset is the demonstration of its sum. As the outside ones are revealed, the displayed running addition will record how much of that sum has been collected.
Watch the five outside ones enter the sum in order. Each addition increases the total by one. The running totals are one, two, three, four, and five. The highlighted source cells show which contributions have already entered. This measured sequence is a way to read the existing relationship, rather than a change to the triangle’s values.
Bring the model of five forward. It contains five individual points in a line. Each point represents one accumulated unit, so the model gives the total a simple visual meaning. We are looking at the completed group of five points. The preceding addition showed how five ones supplied that total.
Return to the triangle and move to seven on the same diagonal. Seven outside ones now contribute. The longer sum follows the same rule as the shorter one. Its final contribution comes from the row immediately above the destination; this endpoint matters when we move to the next kind of diagonal.
Accumulate the counting entries
Move inward to six in row four, position two. Its contributions are one, two, and three from the preceding diagonal. Watch those contributions accumulate: first one, then three in total, then six. We are now adding the counting entries themselves, rather than adding another collection of identical ones.
Bring the triangular model of six forward. It has three rows of points: one in the first row, two in the next, and three in the last. The complete arrangement contains six points. The triangular shape gives the addition one plus two plus three a direct spatial reading.
Move to ten in row five, position two, and look at its complete model. A fourth row of four points extends the triangle. The row lengths are one, two, three, and four, totaling ten. The extra four matches the next contribution from the counting diagonal.
Continue to fifteen in row six, position two. Its model has five rows, ending with a row of five points. The preceding ten together with that new five gives fifteen. Successive triangular entries differ by successive counting entries. The shape and the accumulated sum are two readings of the same count.
Connect the long sum to its parents
Return to the highlighted contributors of fifteen. Read one, two, three, four, and five along the source diagonal. They sum to fifteen. Their displayed slant helps you follow the list, but the rule depends on their row and position: these are the consecutive entries of one fixed diagonal, beginning at its outside one.
Switch to Parents without changing the destination. The two parents are five and ten. The ten already collects the earlier one, two, three, and four. Adding the next five produces fifteen. This shows how a long accumulated sum can be built from the same local addition you studied in the first lesson.
Move down to twenty-one in row seven, position two. Its parents are six and fifteen. The existing total fifteen joins the next contribution six, giving twenty-one. Following this neighboring pair explains the new total without requiring us to start the complete addition again from the beginning.
Return to Diagonals at twenty-one. The full contribution list is now one through six. The local parent picture and this longer highlighted list describe the same entry. One emphasizes the latest step; the other records all the contributions collected so far. We will use that same connection with triangular contributions.
Accumulate triangles into a new diagonal
Move inward again to ten in row five, position three. Its contributions are the first triangular entries: one, three, and six. Watch the running sum become one, then four, then ten. Although this total is ten, we have reached it through a different diagonal relationship from the earlier sum of one, two, three, and four.
Continue to twenty in row six, position three. The contribution list gains the next triangular entry, ten. The accumulated sum is one plus three plus six plus ten, giving twenty. The old total ten and the new triangular contribution ten happen to be equal here, but they have different roles.
Move down to thirty-five. Add the next triangular contribution, fifteen, to the preceding total twenty. The complete sum is one, three, six, ten, and fifteen, totaling thirty-five. Each successive result on this diagonal collects one more triangular entry.
Bring the model of thirty-five forward. It contains five triangular layers of points. Their sizes are fifteen, ten, six, three, and one, from the largest layer to the smallest. Together they contain thirty-five points. These are the same triangular contributions we just added, arranged as a tetrahedral collection.
Look around the layered model
Turn the model to see its depth. The points belong to five triangular layers, even when some overlap in the screen image. Rotation changes which points are easy to distinguish. It preserves the collection itself and the total thirty-five. Use the gaps and the changing overlaps to follow how the layers sit above one another.
Continue around the same model. The smallest layer contains one point, followed by layers of three, six, ten, and fifteen. We are counting points in these layers, not measuring the volume of a filled solid. The model supplies a geometric interpretation of the accumulated triangular entries.
Compare with the preceding tetrahedral entry, twenty. Its complete model has four triangular layers, containing one, three, six, and ten points. The layer of fifteen is absent from this smaller example. The difference between twenty and thirty-five is precisely that next triangular contribution.
Now visit fifty-six, the next entry after thirty-five. This model has six triangular layers, with twenty-one points in the largest layer. The earlier thirty-five and the new twenty-one give fifty-six. The same accumulated-sum pattern continues when the model becomes larger.
The operation continues inward
Return to the triangle at fifty-six. Its highlighted contributions are one, three, six, ten, fifteen, and twenty-one. The model’s layers correspond to these entries. We can move between the geometric collection and the diagonal sum while keeping the same total and the same individual contributions.
Move one diagonal farther inward to seventy in row eight, position four. Its contributors are one, four, ten, twenty, and thirty-five: entries from the tetrahedral diagonal. Their sum gives seventy. The arithmetic operation continues even when we do not display a higher-dimensional point model beside it.
Switch to Parents at seventy. The two parents are thirty-five and thirty-five. One is the accumulated total from the preceding row; the other is the next tetrahedral contribution. Their sum is seventy. Once again, the long diagonal addition advances by a single use of the local parent rule.
Finish with the diagonal sum visible. We began with ones, collected them into counting entries, collected those into triangular entries, and collected triangles into tetrahedral entries. Each level uses the preceding diagonal as its source. In the next lesson, the Hockey Stick view will isolate a chosen run and its destination, so you can read this accumulated-sum relationship anywhere in the triangle.