Mixed Problems: Set 2 (Sums of Coefficients)

Transcript

0:00Introduction

Hello everyone! Okay, one small challenge before we start. Ten C zero times ten C two, plus ten C one times ten C three, and so on, up to ten C eight times ten C ten. Nine products. The first is forty five, the next is one thousand two hundred, and the middle one is forty four thousand one hundred! Will you multiply all nine pairs, and then add them up? That could take ten minutes, right? By the end of this class, you will write the whole sum as one single n C r, in one line. This is Set two of our mixed problems, and sums like these come up often in JEE Main. So, let's begin.

0:43Set 2: sums of coefficients

So, here is what Set two holds. Fifteen solved problems, all in the JEE Main pattern. Okay? Plain sums of the coefficients, weights like r or r squared on each coefficient, and squares and products. All mixed, just like the real exam. Right? Each one should take you one to two minutes. So attempt the whole set first, and only then read the solutions. With me?

1:09Tool 1: put x=1 and x=-1

Tool number one, from Chapter four. Start with one plus x, in round brackets, to the power n, where n is a positive integer. It equals C zero, plus C one times x, and so on, the three dots, up to C n times x to the power n. Here C r, the capital C with a small r below, means n C r, with the small n at the top left. Okay? This is true for every value of x, so we choose a clever one. Which value of x leaves just the coefficients, with nothing attached? x equals one, yes! 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, ending with minus one to the power n, times C n. The left side becomes one minus one, to the power n: zero. Clear? So the C's with an even number below add up to the same as the ones with an odd number below. Together they make two to the power n, so each is half: two to the power, n minus one. Got it?

2:29Tool 2: absorb the weight r

Tool number two, also from Chapter four: absorption. r times C r equals n times n minus one C r minus one: the n minus one sits small at the top left, and r minus one at the bottom right. The weight r gets absorbed, and n comes out in front. With me? Now add over every r. What do all these n minus one C r minus ones add up to? Two to the power, n minus one, exactly! That is tool one again, for n minus one. The big Greek letter sigma means add over all r. So the sum of r C r is n times two to the power, n minus one. Write this down. Okay?

3:12Tool 3: products, look at the indices

Tool number three, for products of coefficients. One plus x to the power m, times one plus x to the power n, equals one plus x to the power, m plus n, where m and n are positive integers. With me? Now compare the coefficients of x to the power k. On the left, x to the power i from the first bracket meets x to the power, k minus i, from the second. So the sum over i, the sigma with a small i below, of m C i times n C, k minus i, equals m plus n, C, k. This is Vandermonde's identity. Clear? So here is the topper's trick: in a sum of products, look at the indices below the C's first. If they add up to the same number in every product, and every possible pair is there, the whole sum is just one coefficient. Right? But what if, instead, the indices differ by a constant? Use symmetry! Replace C j by C, n minus j; the two are equal. So the fixed difference turns into a fixed sum, and we are back to Vandermonde. Got it?

4:20Example 1: sum of r nCr when all the nCr add to 1024

Now, the first problem from the set, an easy one. The sum of n C r, with sigma running from r equals zero below to n on top, is one thousand and twenty four. Find the sum of r times n C r. Pause the video, and try it first. Done? Okay, let's check. By tool one, all the n C r add up to two to the power n. So two to the power n is one thousand and twenty four. What is n? Ten, yes! Two to the power ten is one thousand and twenty four. The double arrow means, so: n equals ten. Okay? Now tool two: n times a power of two. We already know two to the power n is one thousand and twenty four. So the answer is ten times one thousand and twenty four; the dot means times. Right? No, no, careful! The power is n minus one, not n. So the sum of r times ten C r is ten times two to the power nine. Clear? Two to the power nine is five hundred and twelve. So, ten times five hundred and twelve: five thousand one hundred and twenty. Perfect!

5:37Example 2: 10C0 10C2 + 10C1 10C3 + ... as one nCr

Now, remember our challenge from the start? Let's finish it. Ten C zero times ten C two, plus ten C one times ten C three, and so on, up to ten C eight times ten C ten. Pause the video, and try it. Hint: look at the indices. Done? Okay, let's check. In the first product, the indices below are zero and two, and they add up to two. So by Vandermonde, the answer is twenty C two, right? No, no, careful! Look at all the index pairs: zero and two, one and three, and so on, up to eight and ten. Their sums keep changing. But they differ by two, every time. Okay? So, tool three: use symmetry. Each product looks like ten C r, times ten C, r plus two. By symmetry, ten C, r plus two, equals which coefficient? Ten C, eight minus r, exactly! Because ten, take away r plus two, leaves eight minus r. Now the indices are r and eight minus r, and they add up to eight, every time. Very good! So the sum is the coefficient of x to the power eight, in one plus x to the power ten, times one plus x to the power ten. And that product is one plus x, to the power twenty. Right? So the answer is twenty C eight: one hundred and twenty five thousand, nine hundred and seventy. Challenge solved! Nine big products, and one coefficient, in one line. That's it!

7:15Quick revision: x = 1, absorb r, Vandermonde for products

So, quick revision. One: put x equals one, and all the C r add up to two to the power n; the even sum and the odd sum are each two to the power, n minus one. Two: absorb the weight r, and the sum of r C r is n times two to the power, n minus one, not n times two to the power n. Three: for products, make the indices add up to a constant, then read off one coefficient. Now go and attempt the whole set, then the practice problems on this page. You will do well, I promise. All the best!