Overview

Transcript

0:00Introduction

Hello everyone! Okay, a quick challenge before we start. Take one plus x, to the power ten, and write out all its coefficients: one, ten, forty five, one hundred and twenty, and so on, up to two hundred and fifty two in the middle, and back down to one. Eleven numbers in all. Can you add them all up in five seconds, without a calculator? Sounds hard, right? By the end of this video, it takes one line. This chapter is all about such sums of binomial coefficients, and they come up often in JEE, so let's begin.

0:37One notation for the whole chapter

So, the whole chapter starts from one expansion. One plus x, in round brackets, to the power n, where n is a positive integer, as written at the end of the line. It equals C zero, plus C one times x, plus C two times x squared, and so on: the three dots mean and so on, up to C n times x to the power n. Okay? Now look at these coefficients: a capital C, with a small number written below it. So what is C r, with a small r below, short for? Yes, exactly! It is n C r, with the small n sitting at the top left of the capital C, okay? Throughout this chapter, C r stands for n C r: the coefficient of x to the power r in this expansion. We drop the n only to save writing; it is still there. Clear? For example, take n equals ten, from our challenge. Then C zero is one, C one is ten, C two is forty five, and so on, up to C ten, which is one again. Our challenge is to add all of these. With me?

1:40The big idea: put convenient values of x

Now the one big idea of this chapter. This expansion is true for every value of x. So, put in convenient values of x. The big Greek letter sigma means: add up all the terms, C r times x to the power r. Got it? So tell me: which value of x turns every power of x into one, and leaves just the coefficients? Yes, x equals one, correct! Every power of one is one. So C zero plus C one plus C two, and so on up to C n, equals one plus one, to the power n: two to the power n. Right? Now put x equals minus one. The signs alternate: C zero, minus C one, plus C two, and so on, and the last term is minus one to the power n, times C n. The left side becomes one minus one, to the power n, which is zero. Okay? Now add these two results. The odd ones cancel, so twice the even sum is two to the power n. Subtract them, and twice the odd sum is two to the power n. So C zero plus C two plus C four, and so on, equals C one plus C three plus C five, and so on, and each is two to the power, n minus one. Exactly half and half! Clear?

2:57Any polynomial: use f(1) and f(-1)

And this works for any polynomial, f of x, not just one plus x to the power n. Put x equals one, and every term becomes just its coefficient. So the sum of all the coefficients is f of one. With me? Put x equals minus one, and the even powers keep their sign, while the odd powers flip. So the sum of the coefficients of the even powers is f of one plus f of minus one, over two. Right? Then what gives the coefficients of the odd powers? Subtract, yes! f of one minus f of minus one, over two. Keep the order: f of one comes first. Our second example needs exactly this. Got it?

3:39The map of this chapter: one tool per pattern

Now, x equals one and minus one are only the start. Here is the map of the rest of the chapter. When each C r carries a weight r, like the sum of r times C r, we differentiate the expansion. Okay? When each C r is divided by r plus one, we integrate it instead. Clear? For products, like the sum of C r squared, we multiply two expansions together. Right? And to add every second or every third coefficient, we use the roots of unity. Each tool gets its own section. So, look at the pattern first, then pick the tool. With me?

4:14Example 1: sum of 10Cr from r = 0 and from r = 1

Now, remember our challenge from the start? Let's finish it. This example, and the next one, are from the next section, Basic Sums. Find the sum of ten C r, for r from zero to ten: that is the zero below the sigma and the ten on top. Then find the same sum, but starting from r equals one. Pause the video and try it. Done? Okay, let's check. The first sum is all eleven coefficients of one plus x, to the power ten. So put x equals one: two to the power ten, which is one thousand and twenty four. One line, just as promised! Perfect! Now the second sum, starting from r equals one. Same answer, one thousand and twenty four, right? No, no, careful! This is the trap. Starting from r equals one leaves out the very first term, C zero, and C zero is one. Okay? So the second sum is two to the power ten, minus one: one thousand and twenty three. Always check where the sum starts. Got it?

5:20Example 2: sum of odd-power coefficients in (1-2x)^9

Now a harder one. Find the sum of the coefficients of the odd powers of x in one minus two x, in round brackets, to the power nine. Pause the video and try it. Hint: use the f of one and f of minus one rule. Done? Okay, let's check. Call this polynomial f of x: one minus two x, to the power nine. Its coefficients are not plain C r, because of the minus two, but our rule works for any polynomial. Right? First, f of one. One minus two is minus one, and minus one to the power nine is minus one, because nine is odd. Clear? Now f of minus one. What is inside the bracket now? One plus two, yes! Minus two times minus one is plus two. So f of minus one is three to the power nine: nineteen thousand, six hundred and eighty three. With me? Now the odd powers: f of one minus f of minus one, over two. That is minus one, minus nineteen thousand six hundred and eighty three, all over two. Okay? So, minus nineteen thousand six hundred and eighty four, over two: minus nine thousand, eight hundred and forty two. Negative? Yes, and that is fine, because the odd powers here carry minus signs. Just keep the order, f of one first, or you get the wrong sign. Very good!

6:42Quick revision: 2^n, alternating sum 0, f(1) and f(-1)

So, quick revision. One: C r means n C r, the coefficient of x to the power r in one plus x, to the power n. Two: x equals one gives the sum two to the power n, and x equals minus one makes the alternating sum zero. Three: for any polynomial, even powers give f of one plus f of minus one, over two, and odd powers give f of one minus f of minus one, over two. Now open the next section, Basic Sums, and try its practice problems. All the best!