The Method of Differences

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Here is a series: five, plus seven, plus thirteen, plus thirty one, plus eighty five, and so on. Is it an A P? Seven minus five is two, but thirteen minus seven is six. So, no. Is it a G P? Seven over five and thirteen over seven are not equal either. So how do we find the sum of its first n terms? The secret hides in the gaps between the terms. Series like this come up often in JEE, so let's begin!

0:37Look at the gaps

So, take a sequence of terms: capital T one, capital T two, capital T three. The small number below the T tells you which term it is, and the three dots mean and so on, okay? And suppose these terms are neither an A P nor a G P, just like our puzzle. Today's idea is really simple. Don't stare at the terms. Look at the gaps: T two minus T one, T three minus T two, T four minus T three, and so on. We call them the differences. Very often the differences have a clear pattern, even when the terms do not. Clear? Now think. Using only these gaps, how can I climb from the first term, T one, to the n th term? Yes, exactly! Start at T one, and keep adding the gaps. Capital T with a small n below it is the n th term, and T n minus one is the term just before it. So, T one, plus each gap in its brackets, up to the last gap, T n minus T n minus one. See, everything in the middle cancels, and only T n is left. Beautiful, right? Now count. From the first term to the n th term, how many gaps did we add? Only n minus one gaps. Correct! So, in short: T n equals T one, plus sigma, k from one to n minus one, of T k plus one, minus T k. This big symbol is the Greek capital letter sigma, and it means add them all up, with k starting at one and stopping at n minus one. With me?

2:11What the differences tell you

Alright, the most common case. Suppose the differences form an A P, okay? What kind of expression in n will T n be? It's a quadratic in n. Very good! T n is T one plus the sum of n minus one terms of an A P, and that sum is a quadratic in n. So T n equals a n squared, plus b n, plus c. The small two up top means squared, and the double arrow means implies. Here a, b and c are just constants. Right? And if the differences are not an A P? Take the differences of the differences, the second differences. If those second differences form an A P, T n is a cubic. And so on, one degree higher at each level. Got it? Now, what if the differences form a G P, with common ratio r? Then T n looks like a times r to the power n, plus b. The small n up top is the power. Again, a and b are constants, found from the first few terms. With me? And sometimes a polynomial part comes along too, so add one if needed. Remember this form, a times r to the power n, plus b. Example two uses it. Okay?

3:26Example 1: Tn and Sn of 2, 5, 10, 17, 26, ...

Now, Example one, an easy one. Find T n and S n for two, five, ten, seventeen, twenty six, and so on. Capital S with a small n below it is the sum of the first n terms. Pause the video, and try it first. Done? Okay, let's check. The gaps first: three, five, seven, nine. They go up by two each time, so the differences form an A P, with common difference d equal to two. So T n will be a quadratic. Clear? The first difference is three, the second is five, the third is seven. So the k th difference is two k plus one. Okay? Now, T n is the first term, two, plus the sum of these differences. And that sum goes from k equals one up to k equals n, right? No, no, careful! Only n minus one gaps, remember? Go up to n, and you add one gap too many. So, two, plus sigma, k from one to n minus one, of two k plus one. Sigma of two k gives n times n minus one, and the ones give n minus one. Together, n squared minus one. So T n equals n squared plus one. Got it? Now a topper's habit: always check on the first few terms. Put n equals one: one plus one, two. Put n equals two: four plus one, five. Put n equals three: nine plus one, ten. All match. Perfect! Now S n: add T k from k equals one to n, so sigma of k squared, plus sigma of one. From the last lesson, sigma of k squared is n times n plus one times two n plus one, over six. And one, added n times, is just n. So S n equals n, n plus one, two n plus one, over six, plus n. With me?

5:28Example 2: sum of 5 + 7 + 13 + 31 + 85 + ...

Now, Example two. Remember our puzzle from the start? Let's finish it. Find S n for five, plus seven, plus thirteen, plus thirty one, plus eighty five, and so on. Pause the video and try it. Hint: look at the gaps. Done? Okay, let's check. The gaps: two, six, eighteen, fifty four. Each gap is three times the one before, so the differences form a G P, with common ratio r equal to three. That is the secret our puzzle was hiding! Clear? So, T n is five, plus the first n minus one gaps. The k th gap is two times three to the power k minus one; the dot in the middle means times. Their G P sum is two times, three to the power n minus one, minus one, over three minus one. The twos cancel, so T n equals three to the power n minus one, plus four. Right? Quick check. Put n equals one: one plus four, five. Put n equals two: three plus four, seven. Put n equals three: nine plus four, thirteen. It matches. Okay? Now add the terms to get S n. The three to the power k minus one part is a G P, no problem. But look at the plus four in every term. When we add n terms, what does that four become? Four n. Yes, exactly! The four is added once in every term, so n times in all. Many students write just plus four here. Oh no, don't do that! So, S n equals sigma, k from one to n, of three to the power k minus one, plus four n. Got it? That G P has first term one, ratio three, and n terms. Its sum is three to the power n, minus one, over three minus one, that is, over two. So S n equals three to the power n, minus one, all over two, plus four n. Puzzle solved! The gaps were a G P all along. See how neat that was?

7:39Quick revision: differences in AP or GP give Tn, then Sn

So, quick revision. When a series is neither an A P nor a G P, look at the differences. T n equals T one, plus the sum of the first n minus one differences: n minus one, not n! Differences in an A P give a quadratic T n. Differences in a G P give a times r to the power n, plus b. And always check T one, T two and T three, then add up to get S n. Now go and try the practice problems on this page. All the best!