Finding the nth Term of an Exponential Sequence

An old puzzle sets a patch of weed on a pond, and every day that patch covers exactly twice the area it covered the day before. On day thirty the weed has swallowed the whole pond. So the puzzle asks you one thing: on which day was the pond exactly half covered? Day fifteen feels right, since half the days ought to give you half the pond. The real answer is day twenty-nine, with half that water still clear. Multiplying growth hides for weeks and then finishes the job in one step, which is why a rule that adds a fixed amount can never describe it. One note on where this turns up. An nth term written as a power is Cambridge IGCSE Extended material, so if your papers come from another board, the video called Finding the nth Term of a Quadratic Sequence is the one your specification asks for.

First comes the diagnosis, and here are the numbers to diagnose: three, nine, twenty-seven, eighty-one, with nothing else told to you about them. Every other nth term in this topic opens by subtracting each term from the next. Do that here and the gaps come out as six, then eighteen, then fifty-four. Those are not settling down, so go a level lower. The differences of those differences are twelve and thirty-six, growing faster still. No row down there ever turns constant, however far you take it. The gaps are multiplying, because the terms are multiplying. A constant second difference means quadratic, and a constant third difference means cubic. Those methods belong to the videos called Finding the nth Term of a Quadratic Sequence and Finding the nth Term of a Cubic Sequence. A row of differences that keeps growing is the signal to stop subtracting. Try the other operation instead, and divide each term by the one before it. Quick check before I do it. Three things you could try on three, nine, twenty-seven, eighty-one. A, subtract each term from the next. B, divide each term by the one before. C, take the differences twice. Which one gives the same number every time? Take your pick. I'll wait. The answer is B. Nine divided by three is three, twenty-seven divided by nine is three, and eighty-one divided by twenty-seven is three again. That fixed multiplier is called the common ratio, written as r, and a sequence built from one is geometric, or exponential. Subtracting hid the pattern completely, and dividing handed it straight over. That swap is the whole diagnosis.

So to the formula, using the tidiest example in this topic: a population of bacteria that triples every hour. After one hour there are three bacteria, after two hours nine, after three hours twenty-seven, and after four hours eighty-one. You want an expression for the population after n hours. Count the multiplications rather than the terms. After one hour you have multiplied by three once, so the population is three to the power one. After two hours you have multiplied twice, which is three squared. After three hours, three cubed. The power is a tally of how many times you have multiplied. So after n hours the population is three to the power n, and that is the shape every answer in this topic takes: r to the power n. Test it on a term you already have. Three to the power four is three times three times three times three, which is eighty-one, and eighty-one is the fourth term sitting right there. Here is your handle for this video: the ratio goes downstairs, the position goes upstairs. Three is downstairs as the base, and n is upstairs as the power. Now the check that matters most. For that same sequence, three, nine, twenty-seven, eighty-one, which of these is the nth term? A, three n. B, three to the power n. C, n cubed. Have a think. I'll wait. The answer is B, three to the power n. Try A with n as four: three fours are twelve, not eighty-one. Try C: four cubed is sixty-four, which misses as well. A is the one worth naming out loud. Three n means three multiplied by n, and that sequence adds three every step, giving three, six, nine, twelve. Multiplying by three each step builds a power. Adding three each step builds a multiple. The ratio downstairs, the position upstairs, keeps those two apart.

A formula like that earns its keep in two directions, forwards to a term you have never seen, and backwards from a number somebody hands you. Forwards is pure substitution. The population after ten hours is three to the power ten, which comes out as fifty-nine thousand and forty-nine. No chain of terms, no ten rows of arithmetic, just the position dropped into the power. Backwards is the direction exam questions prefer. Is two hundred a term of that sequence? Really that asks whether three to the power n can ever equal two hundred, for a whole number n. So walk the powers: three to the four is eighty-one, three to the five is two hundred and forty-three. Two hundred sits in the gap between them, so no whole number of hours produces it, and it is not a term. Saying that in one sentence is what finishes the answer. Then test any formula against a real term before committing to it. Put n as one, and the first term should come straight back out: three to the power one is three. That test is the safety net for the whole method, since it catches a base read off the wrong pair of terms.

Sequences do not always arrive pure, and the second worked example has had something quietly added on top. The number of new branches at each generation of a growing plant pattern runs three, five, nine, seventeen. Find an expression for the number of new branches at generation n. Try the ratio first, out of habit. Five divided by three is not a whole number, and nine divided by five is not the same value again, so this one is not geometric. So take differences after all. Three to five is two, five to nine is four, and nine to seventeen is eight. Two, four, eight. Those differences are doubling, and doubling differences point straight at the powers of two. Here is why. The powers of two run two, four, eight, sixteen, and the gap between neighbouring powers of two is the smaller one of the pair. So a row of gaps that doubles is the fingerprint of a two to the power n underneath. So write the powers of two underneath, sitting below three, five, nine and seventeen. Then subtract, term by term. That subtraction is yours. Take two from three, four from five, eight from nine, and sixteen from seventeen, then tell me what is left. Pause it there and work it out. I'll wait. One, one, one and one. The same constant every time, so the sequence is the powers of two with one added on, and the nth term is two to the power n, plus one. Check it at generation four. Two to the power four is sixteen, plus one is seventeen, and seventeen is the fourth term. The leftovers will not always be a constant. Had they come out as one, two, three, four, the answer would end in plus n instead. Ratio first, differences second, and when those differences double, subtract the power out and read whatever is left behind.

The wording of these questions does a quiet amount of work, and the phrase to read slowly is in terms of n. That means the answer must be an expression containing the letter n, so line the near-misses up honestly. Eighty-one is a term. Three is the ratio. Multiply by three each time is the rule in words. Only three to the power n is the expression. Which puts the weight on how you write the index. Set the power small and high, clear of the base, so a marker can never read it as a number sitting in front of n. Whichever direction a question runs, the answer on the page shows a power, with n sitting inside it.

The ratio downstairs, the position upstairs. Run back through the method now, from the first suspicion to the finished expression. Differences that never settle, at any level, are the signal. The gaps grow because the terms multiply, so stop subtracting and start dividing. A constant answer from dividing neighbours is the common ratio, r. When that ratio is also the first term, the nth term is simply r to the power n. The power counts how many times you have multiplied, which is why the position sits upstairs and never in front of the base. Use it forwards by substituting a position in, and backwards by walking the powers until you land on the number or step over it. And if the ratio test fails while the differences double, subtract those powers off and read the constant or linear part left behind. You have got this when four multiplying terms hand you an expression in r to the power n, and you check it against the fourth term.

Next in the chapter: Sum of an Arithmetic Series, where a question stops asking for one term and wants the total of a whole run of them.

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