The Numerically Greatest Term

Transcript

0:00Introduction

Hello everyone! Okay, a quick puzzle. Take three plus two x, to the power ten, at x equals one. Eleven terms. Which is the biggest? Last lesson, the middle binomial coefficient won. So the middle term, T six, right? Hmm, not so fast. Today, one small number called m will pick the winner every time. This idea is asked in JEE, so let's begin.

0:28Compare each term with the one before it

So, here is our puzzle as a bar chart: the eleven terms, T one to T eleven. The bars rise, reach a peak, and then fall. And the tallest bar, in orange, is T five, not T six! Its value, written on top, is ten C four, times three to the power six, times two to the power four: about two point four five million, at the dashed line. Okay? Now think. How can you tell that the next bar will be taller? What should you compare? Compare each term with the one before it, exactly! First, the general term of x plus a, to the power n, with n a positive integer. T r plus one, the capital T with r plus one below, equals n C r, with the small n on the top left of C, times x to the power n minus r, times a to the power r. And r runs zero, one, and so on up to n, so the counting starts at r equals zero. Right? Now divide T r plus one by T r. The vertical bars mean the modulus, the size without the sign, because we want the numerically greatest term. From the general term, this ratio is n minus r plus one, over r, times the modulus of a over x. The dot means times. Clear? So tell me. When is T r plus one at least as big as T r? When this ratio is at least one, yes! Greater than with a line below means greater than or equal to, and the double arrow means if and only if. Multiply across: n minus r plus one, times mod a, is at least r times mod x. Okay? Now bring both r terms to one side. We get r less than or equal to, n plus one times mod a, over, mod x plus mod a. So each term is at least as big as the one before it, as long as r is at most this number. The note beside the chart says it too: the ratio is at least one while r is at most four point four, so the terms rise up to T five, then fall. With me?

2:40The rule: one number m decides

So let's call this number m. m equals n plus one, times mod a, over, mod x plus mod a. Here x and a stand for the two terms of the bracket as actual numbers. So if the question has a variable in it, put its value in first, then take the sizes. Got it? Try it on our chart. n is ten, the first term is three, and two x is two at x equals one. So m is eleven times two, over three plus two: four point four. Right? Now careful. m is four point four. So the greatest term is T four, isn't it? No, no, careful! The orange bar is T five. The ratio is at least one for r up to four, and r equals four is the climb from T four to T five. So when m is not an integer, the numerically greatest term is T of, the integer part of m, plus one. The square brackets around m mean the greatest integer function, the integer part of m. Okay? The integer part of four point four is four, plus one is five: T five, the orange bar! Forgetting this plus one is a very common slip, so write it down. Clear? And if m is exactly an integer? Then at r equals m the ratio is exactly one, so T m plus one has the same size as T m. T m and T m plus one are numerically equal, and both are the greatest. We will see this in Example two. Got it?

4:12Example 1: greatest term of (4-3x)^7 at x = 2/3

Now Example one. Find the numerically greatest term of four minus three x, to the power seven, when x equals two over three. Pause the video and try it. Done? Okay, let's check. Here n is seven, and the first term is four. Now the second term. Is mod a equal to three? Oh no, careful! Not three. The second term is minus three x, and x is two over three. So minus three times two over three is minus two, and mod a is two. So, put the value of x in first. With me? Now m. Seven plus one is eight, times two, over four plus two. Sixteen over six, which is eight over three. The wavy equals sign means approximately: about two point six seven. Right? Not an integer. So the integer part of m is two, plus one is three: T three. Not T two! Remember the plus one. Okay? Now the value. T three is seven C two, times four to the power five, times minus two, squared. That is twenty one, times one thousand and twenty four, times four: eighty six thousand and sixteen. Positive, because the power of minus two is even. Perfect!

5:38Example 2: (3-5x)^15 at x = 1/5, two equal greatest terms

Now Example two, a little harder. Find the numerically greatest term of three minus five x, to the power fifteen, when x equals one over five. Pause the video and try it. Hint: put the value of x in first. Done? Okay, let's check. Minus five x, at x equals one over five, is minus one. So the general term is fifteen C r, times three to the power fifteen minus r, times minus one to the power r. Right? Now m. Fifteen plus one is sixteen, times one, over three plus one. Sixteen over four is four. Exactly four, an integer! Clear? So which term is the greatest now? T four? T five? Or both? Both, yes! At r equals four the ratio is exactly one, so T five has the same size as T four. Both are the greatest. Write only one of them, and the answer is incomplete. Got it? Now the sizes. Mod T four is fifteen C three, times three to the power twelve. That is four hundred and fifty five, times five hundred and thirty one thousand, four hundred and forty one. Okay? And mod T five is fifteen C four, times three to the power eleven: one thousand three hundred and sixty five, times one hundred and seventy seven thousand, one hundred and forty seven. See the shortcut? One thousand three hundred and sixty five is three times four hundred and fifty five, and three to the power twelve is three times three to the power eleven. So the products are equal. Very good! Both come to two hundred and forty one million, eight hundred and five thousand, six hundred and fifty five. Now the signs. T four has r equals three, an odd power of minus one, so T four is less than zero, which is less than T five. In size they tie; that is why the word numerically matters. With me?

7:41Quick revision: the number m and when two terms tie

So, quick revision. One: the terms grow while the ratio of neighbours, in modulus, is at least one. Two: m is n plus one times mod a, over mod x plus mod a. If it is not an integer, the greatest term is T of the integer part of m, plus one. Three: if m is an integer, T m and T m plus one tie as the greatest. And if you forget m, just write the ratio of neighbours and solve; the rule comes right back. Now try the practice problems on this page. All the best!