Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take a positive integer n, and all its binomial coefficients: n C zero, n C one, and so on, up to n C n. First, add up one over each of them. Then, add up r over n C r, for r from zero to n. Now, what is the second sum divided by the first? Hmm, adding up either sum directly looks hopeless, right? And yet the answer is really simple, and today you will get it in four short lines. This is the unit problem bank for the Binomial Theorem, where problems from the whole unit come mixed together, just like a real JEE paper. So first, let me show you how to use it well. Let's go.

0:45How this chapter works

So, first, how this chapter is built. Part A, at JEE Main level: twenty solved problems, from the whole unit, all mixed together. Okay? Now, a question for you. When should you open Part A? Right after Chapter one, while the topic is fresh? No! Only after Chapters one to seven. The little dash means one to seven. And attempt it as a forty minute test, with a clock running. Right? So, twenty problems in forty minutes. On average, how many minutes does each problem get? Forty divided by twenty: two minutes per problem, on average. That sign is the division sign, okay? Two minutes is tight, so keep moving. Clear? Next comes Part B: twelve longer problems, at JEE Advanced level. With me? Then Part C, the concept checks. These are the statements tested most often, in true or false form, or in which is correct form. Okay? And at the very end, a unit test of fourteen problems, answers only, no worked solutions. There, you check yourself honestly. Got it?

1:53Part B: four ideas to look for

Alright, now Part B, at JEE Advanced level. Twelve problems: two medium and ten hard, so give each one real time. Right? They combine several ideas. First, sums with integration, for terms like n C r over r plus one. Here n C r is the C with a small n at the top left, and r at the bottom right. Okay? Second, pairing tricks: replace r by n minus r, so each term meets its partner from the other end. Our puzzle from the start is exactly this type. Got it? Third, conjugate surds. Pair root two plus one, to the power n, with its partner, root two minus one, to the power n. The tick-shaped sign is the square root. With me? And fourth, finite differences, like this one. The sum, for r from zero to n, of minus one to the power r, times n C r, times r to the power n, equals minus one to the power n, times n factorial. The big Greek letter sigma means add up, and the exclamation mark is the factorial. Clear?

2:59Part C: is the statement exactly right?

Now, two items from Part C, okay? First: the sum of r times n C r, for r from zero to n, equals n times two to the power n. The question mark asks: true or false? Oh no, careful! It is false. The correct sum is n times two to the power n minus one. Students remember the n in front, and forget the minus one in the power. One small slip, and the answer doubles! That is exactly what Part C catches. Right? Second item. One plus x, to the power one half, equals one, plus x over two, minus x squared over eight, and so on; the dots mean the series goes on forever. Can we use it at x equals three? No! The index here is one half, a non-integral index, so this series is valid only when mod x is less than one. The two straight bars mean mod, the size of x, and that sign means less than. At x equals three, we cannot use it at all. Got it?

3:58Example 1: term independent of x in (2x^2 - 1/x)^6

Now, the first example. It is the very first problem of Part A, marked easy. Find the term independent of x in two x squared minus one over x, the whole thing to the power six. Independent of x means the term with no x at all. Start your clock, pause the video, and try it. Done? Okay, let's check. Always start with the general term, T r plus one: capital T, with r plus one written small at the bottom right. T r plus one equals six C r, times two x squared to the power six minus r, times minus one over x, in brackets, to the power r. And r is counted from zero: zero, one, and so on up to six. With me? Now tell me. What power of x does this term carry? Two x squared, to the power six minus r, gives x to the power two times six minus r. And one over x, to the power r, gives x to the power minus r. Together: twelve minus three r. Right? For no x at all, the power must be zero. Twelve minus three r equals zero, so r equals four. Clear? So the answer is six C four, which is fifteen, right? No, no, wait! Six C four is only the binomial coefficient. The numbers inside the brackets multiply it. Two to the power six minus four, that is two squared, and minus one to the power four, which is plus one, since the power is even. The dots mean times. So fifteen times four: sixty. And notice, r equals four gives T five, the fifth term, not the fourth. Okay?

5:38Example 2: sum of r/nCr over sum of 1/nCr by pairing

Now, our puzzle from the start. It is a Part B problem, marked hard. S n, capital S with a small n at the bottom, is the sum of one over n C r, for r from zero to n. And small t n is the sum of r over n C r. Find t n over S n. Pause the video, and try it. Done? Okay, let's check. Adding up these sums directly is hopeless. So, which of the four Part B ideas fits here? The pairing trick, yes, exactly! Replace r by n minus r. As r runs from zero to n, n minus r runs over the same values, backwards. And n C n minus r equals n C r, so the denominators do not change. Same terms, in reverse order. Okay? So, in t n, the top r becomes n minus r, and the bottom stays n C r. Now split the fraction: n times the sum of one over n C r, minus the sum of r over n C r. With me? Now look at those two sums carefully. Have you seen them before? Yes! The first one is S n, and the second one is t n itself. So t n equals n S n, minus t n. Right? Now take t n across: two t n equals n S n. So t n over S n is just n over two. That's it! Puzzle solved! And a quick check, for n equals two. S two is one, plus a half, plus one: five over two. And t two is zero, plus a half, plus two: also five over two. Their ratio is one, and two over two is one. Perfect! Clear?

7:19Quick revision: how to use Parts A, B and C

So, quick revision. Part A: a forty minute test, after Chapters one to seven. Part B: spot the idea: integration, pairing, conjugate surds, or finite differences. Part C: is the statement exactly right? Then, the unit test. Now open Part A, start your clock, and begin. Then try the unit test at the end of this chapter. All the best!