Overview

Transcript

0:00Introduction

Hello everyone! Okay, a quick challenge before we start. Take three to the power one hundred and one. That is a forty nine digit number. Now divide it by twenty six. What is the remainder? Long division with forty nine digits? That would take forever, right? In this chapter, a question like this should take you one to two minutes, no more. The binomial theorem comes up often in JEE Main, so this speed really matters. Today, I will show you how this chapter of mixed problems works, and how to use it well. Let's go.

0:37Why this chapter

So, first, why this chapter matters. The binomial theorem gives one or two questions in many JEE Main shifts. So this unit is really worth your time, okay? Now, the real JEE Main questions, from twenty twenty one to twenty twenty six, are already at the end of Chapters one to six. The short dash on screen, between the years and between the chapter numbers, just means to. You meet those questions there, chapter by chapter. Right? And this chapter is different. Its problems are original, written in the same style as those real ones. So everything here is fresh practice, questions you have not seen before. Got it?

1:21Six question types

Alright, now think back over the six chapters of this unit. Which types of binomial questions do you expect in the exam? Here they are, six types. First, find a coefficient. Or find the term independent of x, that is, the constant term, the term with no x in it at all. Okay? Next, a condition on consecutive coefficients, that is, coefficients that sit next to each other in the expansion. And a sum of binomial coefficients. Clear? Then, a count of rational terms. The greatest term. And finally, a remainder, just like our challenge from the start. You will meet all six types in this chapter. With me?

2:04Three sets, one clock

Now, the problems come in three sets. Set one: terms and coefficients. Set two: sums of coefficients. Right? And Set three: greatest terms, remainders, and multinomials, that is, expansions with more than two terms inside the bracket. Okay? Each set has fifteen solved problems, and each problem should take you one to two minutes. So tell me, how long should one whole set take? Fifteen to thirty minutes. Correct! Fifteen times one, and fifteen times two. So start a clock before you begin, and see where you stand. Clear? And the most important rule: attempt the whole set before reading any solution. Peeking at each answer feels comfortable. But it hides your real speed, and your real mistakes. So finish the set first, then check. Got it?

2:56Example 1: coefficient of x^11 in (x^2 + 2/x)^10

Now, our first example. It is from Set one, and it is marked easy. Find the coefficient of x to the power eleven in x squared plus two over x, all inside the big round brackets, raised to the power ten. The small numbers at the top right are the powers, okay? Start your clock, pause the video, and try it. Done? Okay, let's check. Always start with the general term. Capital T, with r plus one written small at its bottom right, is the term number r plus one, with r counted from zero. For a positive integer power, here ten, it is ten C r, with the small ten at the top left of the capital C, times x squared to the power ten minus r, times two over x, to the power r. Right? Now a topper's trick: collect the power of x first. x squared to the power ten minus r gives x to the power twenty minus two r. And one over x to the power r takes away r more. So x to the power twenty minus three r. The two just stays as two to the power r, a plain number. Clear? We want x to the power eleven. So, which r do we need? Twenty minus three r equals eleven. The double arrow means this gives: three r is nine, so r equals three. Yes, exactly! Only one value of r works here. With me? Now ten C three is one hundred and twenty. So the coefficient is one hundred and twenty, right? No, no, careful! That is the trap. The two inside two over x is also raised to the power r. So we must keep two cubed, that is eight. The dot means times. One hundred and twenty times eight: nine hundred and sixty! Under a minute, and done. Perfect!

4:52Example 2: remainder when 3^101 is divided by 26

Now, remember our challenge from the start? Let's finish it. It is from Set three, and it is marked medium. Find the remainder when three to the power one hundred and one is divided by twenty six. Start your clock, pause the video, and try it. Done? Okay, let's check. Here is the trick for remainders. Find a small power of three that sits just one above a multiple of twenty six. Which power? Three cubed! Yes, exactly. Three cubed is twenty seven, and twenty seven is twenty six plus one. Okay? Now, ninety nine is three times thirty three. So three to the power ninety nine is three cubed, in brackets, to the power thirty three. That is twenty six plus one, to the power thirty three. Thirty three is a positive integer, so expand it with the binomial theorem. Every term carries a factor of twenty six, except the last one, which is just one. So it is twenty six k plus one, where k is a whole number. Clear? So the remainder is one, right? No, no, wait! That is the remainder for three to the power ninety nine. We need three to the power one hundred and one, so two more threes. So write it as three squared, times three to the power ninety nine. So we get nine times, twenty six k plus one. Got it? Multiply out: twenty six times nine k, plus nine. Nine is less than twenty six, so the remainder is nine! See, a forty nine digit number, and we never had to write it down. That's it!

6:32Quick revision: timed sets, coefficients, remainders

So, quick revision. Three sets of fifteen problems: one to two minutes each, and the whole set before any solution. For coefficients, find the power of x in T r plus one, the general term, first. Then keep every number. And for remainders, write a power of the base as a multiple of the divisor, plus one. Now open Set one, start your clock, and begin. You are ready for this, I promise. All the best!