Proving the expansion for every positive power by induction

0:00There are rows of coefficients now, three regularities that hold every time you look, and a formula that matches all of them. It would be easy to think there is nothing left to do. There is. Everything so far is a check, and every check is finite. The rows for the first, second, third and fourth powers were multiplied out and confirmed, which is fourteen numbers altogether. The row for the fifth power was used and never checked.

0:33And the zeroth power is not even a candidate, because the claim we are about to make is about positive whole powers, and the zeroth line came with a condition attached that none of the others carry. So: fourteen confirmed numbers, and a claim about an unending list of rows. The gap between those two things is not a technicality. It is the entire reason the next argument exists. To see how wide that gap is, here is a rule I made up.

1:09At the two ends of every row it says one. Everywhere in between it takes the true entry and adds the product of n minus one, n minus two, n minus three, n minus four, and n minus five. Below the sixth power that product is nothing at all. So this rule agrees with the truth at every entry of every row anybody checked. It has a one at both ends of every row.

1:38Every row of it reads the same in both directions. It passes every test the pattern was ever put to. And at the sixth power it adds a hundred and twenty to each of the five interior entries, giving one, a hundred and twenty-six, a hundred and thirty-five, a hundred and forty, and back down the other side. Exactly one thing catches it, and that is multiplying the bracket in.

2:06Out to the eighth power the rival breaks the addition rule at eighteen of the twenty-eight interior places, the first of them at the sixth power. So what would close the gap? A principle, and it is worth stating carefully, because it is doing all of the work. If a claim holds for a first case, and if, wherever it holds, it also holds one step further along, then it holds for every case from the first onwards.

2:39That is the whole of it. Two things: a starting point, and a step you can take from anywhere. Neither one of them alone gets you anything. And notice what the step is not. It is not the observation that the claim held here, and here, and here as well. It is a single argument that works at an unnamed place, at n for any n, which is exactly why one step can cover an unending list.

3:09So where does the starting point go? At the first power, not the zeroth. Put the first power in and the row has two entries, both of them one, so the claim says the bracket equals the first thing plus the second thing. Which it does. That is the base case, and it is done. The zeroth power is left out on purpose. The zeroth power of a bracket is one, unless the bracket comes to nothing, and nought to the nought names nothing at all.

3:42That condition is real. It is why the zeroth line always came with a caveat, and it is why the claim is made about positive powers and starts where it does. Now watch what happens if you drop one half of the principle. Here is a statement about these same rows: the row for index n adds up to three times two to the n. Test the step on it.

4:11A row gets copied twice by the bracket, so whatever a row adds to, the row below it adds to twice that, and out to the twelfth power that never once fails. Three times two to the n, doubled, is three times two to the n plus one. So the step holds. It holds every single time. And the statement is false at every index there is, because the first row adds to two and the statement says six.

4:42There is no base case, so the step is carrying nothing forward at all. It never gets started, and it reaches nought. Now the other way round. Say instead that the row for index n adds up to n plus one. At the first power that is two, and the first row does add to two, so the base case here is perfectly fine. Now try the step. Doubling two gives four, and the statement wants three.

5:13The step fails immediately. So that claim reaches exactly one case, the one you checked by hand, and nothing after it. The true statement, that the row adds to two to the n, passes both halves, and the machine runs it out to the twelfth power without stopping. Two failures, two different halves, and you need both halves every time. So to the step itself, which is where the mathematics of this actually is.

5:45Assume the claim at one index. Not at every index, and not at the indices already checked, but at one unnamed index n, and nothing else whatever. Write that row out. It has n plus one terms. The first letter's index falls from n down to nought, the second climbs from nought up to n, and the coefficient at each place is the selection count for that row and that place, carried as a symbol and not worked out.

6:16That assumption is the only thing on the table. Everything that follows has to come out of it. Multiply that row by one more bracket. The bracket has two terms, so it distributes: the first letter multiplies every term of the assumed row, and so does the second. Take the fifth power as the picture. Six terms go in and twelve come out, and not one coefficient has changed. The first copy has every first index bumped by one.

6:51The second copy has every second index bumped by one. The numbers in front were never touched at all. That is the whole of the multiplication: two full copies of the row you assumed, the same list of coefficients twice over, differing only in which index went up. Now line the two copies up. Read a place along the new row by its second index. The first copy's terms sit at places nought through five.

7:24The second copy's terms sit at one through six, which is the same list slid one place to the right. So at the two outside places only one copy arrives, and at every place between them two coefficients land together. Gather them and twelve terms become seven. Now look hard at what is standing at, say, place three. It is not a number, and it is not yet the coefficient the claim wants.

7:53It is two selection counts side by side: the one at place three of the assumed row, and the one at place two. That is all the multiplication gives you. So the last thing needed is whatever turns two adjacent counts into the single one below them. And that is an identity about choosing, not about brackets. Take the selections of three from eight and single one bracket out. Twenty-one of them name it and thirty-five do not, nothing is in both families, nothing is left out, and the two numbers add to fifty-six, which is the count you wanted.

8:35That argument is not made here. It is imported from counting, and this is the one thing the proof borrows. Three small facts finish the step: the count at the start of the new row is one, the count at the end is one, and two adjacent counts fuse into the one below. Withhold each of the three in turn and count what goes unjustified. Across the twelve steps there are a hundred and two lines to justify.

9:07The fact about the start carries twelve of them, the fact about the end carries twelve, and the fusing identity carries seventy-eight. The one doing the work is the one that was borrowed. Base case, plus step, and the principle hands you every positive whole power. Not four rows and not six, but every one of them, without ever writing another out. And be exact about what it does not hand you.

9:37The claim was made about positive whole powers, so that is what it covers. Half powers, negative powers, powers that are not whole numbers at all: nothing above says a single thing about any of them. The selection count itself refuses at a place past the end of its row, and at an index below nothing, so the machinery does not even pretend to reach there. Those cases are real mathematics.

10:07They are simply not this. With the theorem proved, the pattern-spotting from before can be restated as consequences of it. A row for index n has n plus one terms. The first index falls, the second climbs, and in every term the two of them add to n. And the whole statement compresses to a single line: a sum, with a counter running from nought up to n, of the selection count times the first letter to the n minus k, times the second letter to the k.

10:44That short form only reproduces the long one because a zeroth power of a letter is read as one. Withdraw that convention and, across the rows from one to twelve, twenty-four of the ninety terms cannot be written down at all. And every one of those twenty-four is at an end of its row. Finally, what the theorem actually hands you. Take the bracket built from a letter and two, raised to the sixth.

11:15Row six is one, six, fifteen, twenty, fifteen, six, one, and each entry gets multiplied by the matching power of two. That gives one, twelve, sixty, a hundred and sixty, two hundred and forty, a hundred and ninety-two, and sixty-four. Look at the fifth of those. The selection count behind it is fifteen and the coefficient is two hundred and forty, so the counts and the coefficients part company fast once the terms are not bare letters.

11:50And the terms need not be whole. Take a letter squared, plus three divided by that same letter, raised to the fourth. You get the letter to the eighth, twelve of it to the fifth, fifty-four of it squared, a hundred and eight over it, and eighty-one over its fourth power. That expansion has nothing to say when the letter is nought, because one of the terms it was built from does not exist there.