Subtracting a scaled copy of the total to collapse it to two terms

0:00Here is a list that multiplies by two thirds at every step. One, two thirds, four ninths, eight twenty-sevenths, and on. Add the first five of them. You can. It is five additions and you will get the right number. Now add the first fifty. You can do that too, and it is only forty-nine more additions, but something has gone wrong with the question. A total you can only reach by doing every addition is not really an answer. It is a promise that you did the work.

0:35What we want is a way to write the total that has no dots in it at all — no "and so on", no "keep going". And the thing worth knowing is that there is exactly one manoeuvre behind it, and it is short enough to rebuild from nothing every time you need it. So do not memorise what comes next. Watch what it is made of. Write the total down and give it a name. Call it S.

1:03S is the opening term, plus the opening term times the ratio, plus the opening term times the ratio squared, and so on to the last one. Now here is the whole idea, and it is one line long. Write the same total again underneath, multiplied through by the ratio. Every term on the first line gets multiplied by r, so the second line is r times S. That is a legal thing to do — multiplying both sides of a true statement by the same number keeps it true.

1:41But look at what it did to the terms. The opening term became the opening term times r, which is the second term of the first line. The second became the third. The third became the fourth. Every term of the second line is a term of the first line, standing one place further along. So slide the second line one place to the right and set them beside each other.

2:09Now they overlap almost everywhere. The first line runs from position one to position n. The second runs from position two to position n plus one. They share positions two through n — that is n minus one places where both lines have something. And at every one of those shared places, the two lines hold the same term. That last part is worth pausing on, because it is doing the work and it is easy to take for granted.

2:41The lines do not agree because they are the same length. They agree because the second one was scaled by the ratio. Scale it by anything else and watch: take a list doubling by three and multiply the copy by five instead. The two lines still share positions two, three, four and five — but the number of those places where the terms actually agree is nought. So the overlap is not bookkeeping. It is the ratio, doing the only thing it can do.

3:17Now subtract the second line from the first. Everything in the shared block cancels, because in that block the two lines are identical. Count what that removes. Each line holds n terms, so there are two n terms written down. The shared block is n minus one terms in each line, so two n minus two of them go. Take a five-term run: ten terms written down, eight cancelled, two left standing.

3:47One survivor from the top line, at position one — that is the opening term, which the second line never reached. One survivor from the bottom line, at position six — that is the opening term times r to the fifth, which ran off the end. Two survivors, every time. Not because the formula says so — because those are the only two positions the overlap misses. Seventy-two of these subtractions are run here, across nine different lists and every run length from two to nine.

4:25The number that left more than two standing, or fewer, is nought. So write down what is left. On the left you have S minus r S, which is S times one minus r. On the right you have the opening term, minus the opening term times r to the n. That is the whole content of the manoeuvre, and it was checked here the honest way. Across all seventy-two runs, the number where subtracting left anything other than one minus r times the honestly added total is nought.

5:03One step remains. Divide both sides by one minus r, and S stands alone. But look hard at that divisor before you use it. One minus r is nothing exactly when r is one. So the last move of the derivation is illegal for one single value of the ratio. That is not a footnote somebody added later. It is a hole the manoeuvre itself dug, and it has to be filled separately.

5:35So take that case on its own terms. Every term is the same as the one before it. The manoeuvre still runs — that is the surprise. Write a six-term flat list, write the scaled copy beneath it, subtract. Ten of the twelve terms still cancel and two still survive, exactly as before. But the two survivors are the same number, so what is left is nothing. And one minus r is also nothing. The manoeuvre has told you that nothing equals nothing.

6:09True, and completely useless — which is why the case is carved out before the division rather than patched afterwards. The honest total there is a count. Five, seven times over, is thirty-five, and it took seven additions to say so. Over every run length from one to eight, the number of times counting disagreed with adding the terms out is nought. Now, you will see this written two ways, and it is worth being clear that it is one formula.

6:44One version has one minus r on the bottom and one minus r to the n on top. The other has r minus one on the bottom and r to the n minus one on top. Take a halving list, five terms: the first top is thirty-one thirty-seconds and the second is minus thirty-one thirty-seconds. The first bottom is a half and the second is minus a half. Both the top and the bottom have been negated, and negating both changes nothing at all.

7:18They are checked here against every case in the file, and the number of times the two forms disagreed with each other is nought. The number of times either of them disagreed with adding the terms out one by one is also nought. So use whichever one keeps your signs comfortable. That is a convenience, not a rule. Three shapes of question use this, and the first is the plain one.

7:48Take one, two thirds, four ninths, and on, and total the first five terms. The opening term is one and the ratio, found by dividing, is two thirds. Two thirds multiplied by itself five times over is thirty-two two hundred and forty-thirds. One less that is two hundred and eleven, over two hundred and forty-three. Divide by one minus two thirds — that is, multiply by three — and you get two hundred and eleven over eighty-one.

8:21Add the five terms out by hand and you get two hundred and eleven over eighty-one. Over run lengths one to eight the number of times those two answers differed is nought. That agreement is the point of doing it: the formula is not a substitute for the addition, it is the same addition with the middle removed. The second shape runs the same machinery the other way. Here is a list opening at three and halving: three, three halves, three quarters, and on.

8:56How many terms does it take to reach a total of three thousand and sixty-nine, over five hundred and twelve? Now the count of terms is the unknown, and it is sitting in an exponent. Written out, the total of n terms is six times the quantity one minus one over two to the n. Set that equal to the target and the whole thing collapses to two to the n equals one thousand and twenty-four.

9:25Ten terms. And nine terms gives three thousand and sixty-six over five hundred and twelve, so the answer really is the tenth. The last step there is matching powers, which is the same move as the previous topic, arriving in a new place. And a target the running total never reaches is refused outright rather than answered — a search that always finds something has not found anything. The third shape is the one that looks like a trick and is not.

10:01Total the list seven, seventy-seven, seven hundred and seventy-seven, and on. Before anything else, check whether it is a progression, because if you skip that step the rest looks like sleight of hand. Seventy-seven over seven is eleven. Seven hundred and seventy-seven over seventy-seven is one hundred and eleven over eleven. Those are different numbers, so the ratio is not constant and this is not a progression. The formula does not apply to it.

10:36So convert it into something the formula does apply to. Seven is seven ninths of nine. Seventy-seven is seven ninths of ninety-nine. And a string of nines is a power of ten less one. So the whole total is seven ninths of a genuine progression of powers of ten, minus n copies of one. Check it at two terms before trusting it at fifty: the expression gives eighty-four, and seven plus seventy-seven is eighty-four.

11:09Across the digits seven and eight, and every run length up to seven, the number of times the converted total disagreed with the plain addition is nought. One more shape, and it is the prettiest. Three terms of a progression multiply to one below nothing and total thirteen twelfths. Find them. The move is to write the three terms centred on the middle one: a over r, then a, then a times r.

11:41Multiply those three together and the ratio cancels itself away, leaving a cubed. So a cubed is minus one, and among all the candidate values the number whose cube is minus one is exactly one of them: minus one itself. Put that back into the total and search for the ratio. Two values fit: minus four thirds and minus three quarters. They give four thirds, minus one, three quarters — and the same three read backwards.

12:14That is one answer written twice, which is what you should expect: nothing in the question said which end to start from. The centring is the whole trick, and it is worth seeing why: putting a in the middle is what makes the product forget the ratio. You will meet questions that attach a small condition to the letters, and they look like fussiness. Total one, minus a, a squared, minus a cubed — with a not equal to minus one.

12:47Total a cubed, a to the fifth, a to the seventh — with a not equal to one or minus one. Those are not arbitrary. The first list has ratio minus a, and among the candidates the only value of a making that ratio one is minus one. The second has ratio a squared, and the only two values making that one are one and minus one. Each exclusion names exactly the value that would make the ratio one — which is the value that makes one minus r nothing.

13:24It is the hole from the derivation, showing up again in a question, wearing different clothes. And there is one more limit worth naming honestly. A list like root seven, root twenty-one, three root seven has a perfectly good ratio of root three — but its terms alternate between two different roots, and the checker behind this video refuses to add them at all, so no total of that kind is claimed here.

13:56Finish with the question this whole idea was invented for. You have two parents. They have four parents between them. Those four have eight. How many ancestors is that over ten generations? Opening term two, ratio two, ten terms. The tenth generation alone is one thousand and twenty-four people. And the total, added out one term at a time, is two thousand and forty-six. Which is two times one thousand and twenty-three, exactly as two lines and a subtraction predict.

14:33Ten additions to do it by hand. One subtraction to do it for any number of generations at all. That is what the manoeuvre buys, and you now own it outright: write the total, write it again scaled by the ratio, line them up, take one from the other, and divide by what is left — remembering that what is left is nothing when the ratio is one.