Coefficients in Sums of Binomials

Transcript

0:00Introduction

Hello everyone! Okay, a small challenge first. One plus x to the power four, plus one plus x to the power five, and so on, up to one plus x to the power ten. Seven brackets. I want one number: the coefficient of x to the power four in the whole sum. Expand all seven? That is fifty six terms! There must be a faster way, right? There is, and it takes one line. Sums like this come up often in JEE, so let's begin.

0:33One coefficient from each bracket

So, here is the general sum: one plus x, in round brackets, to the power m, plus one plus x to the power m plus one, and so on, up to the power n. The three dots mean and so on, and every power is a positive integer. Okay? Quick one. In a single bracket, one plus x to the power k, what is the coefficient of x to the power r? Yes, exactly! It is k C r, with the small k at the top left of a capital C, and r at the bottom right. Right? So in the whole sum, the coefficient of x to the power r is m C r, plus m plus one C r, and so on, up to n C r. With me? But adding them one by one is slow. So, two fast ways: the hockey stick, or sum the G P first; G P means geometric progression.

1:29Way 1: the hockey stick

First, Pascal's triangle. In the figure, the blue arrows show Pascal's rule, and the orange band is a hockey stick: its total sits one row below, one step to the right. In symbols: r C r, plus r plus one C r, and so on up to n C r, equals n plus one C r plus one. Okay? But our sum starts at m C r, not at r C r. So, is the answer still n plus one C r plus one? No, careful! The hockey stick must start on the edge, at r C r. Ours starts later, so the first few terms are missing. Clear? The missing terms are r C r, and so on, up to m minus one C r. That is a hockey stick too, and it adds up to m C r plus one. Right? So, full hockey stick, minus the missing part. The capital Greek letter sigma means add up, from k equals m, below, to n, on top. The sum of k C r, for k from m to n, equals n plus one C r plus one, minus m C r plus one. Got it?

2:35Way 2: sum the G.P. first

Now the second way. Each bracket is the one before, times one plus x. So it is a G P: first term one plus x to the power m, common ratio one plus x. Okay? The G P formula divides by the common ratio minus one, which is just x. So the sum of one plus x to the power k, for k from m to n, equals one plus x to the power n plus one, minus one plus x to the power m, all over x. Got it? Now, for x to the power r in the sum, we look for x to the power r in the numerator, right? Oh no, careful! Dividing by x lowers every power by one. So we need x to the power r plus one in the numerator. Clear? And there, its coefficient is n plus one C r plus one, minus m C r plus one. The same answer as the hockey stick! With me?

3:31One more G.P. that collapses

One more closed form. The sum of x to the power k, times one plus x to the power n minus k, for k from zero to n. Written out: one plus x to the power n, plus x times one plus x to the power n minus one, and so on, up to x to the power n. The power of x goes up as the power of one plus x goes down. Okay? Sum it as a G P, and it collapses to just two terms: one plus x to the power n plus one, minus x to the power n plus one. Beautiful, right? Check n equals one. Left side: one plus x, plus x, that is one plus two x. Right side: one plus x, the whole squared, minus x squared: again one plus two x. Perfect!

4:20Example 1: x^4 in (1+x)^4 + (1+x)^5 + ... + (1+x)^10

Now, back to our challenge. Find the coefficient of x to the power four in one plus x to the power four, plus one plus x to the power five, and so on, up to one plus x to the power ten. Pause the video and try it. Done? Okay, let's check. From each bracket: four C four, plus five C four, and so on, up to ten C four. Okay? It starts at four C four, and r is four: right on the edge. So it is a full hockey stick. With me? The last term is ten C four, so the answer is eleven C four, right? No, no, wait! Both numbers go up by one: eleven C five. The small dots mean times. Eleven times ten times nine times eight times seven, over five times four times three times two times one: four hundred and sixty two. Clear? To be sure, add the seven numbers: one, five, fifteen, thirty five, seventy, one hundred and twenty six, two hundred and ten. Four hundred and sixty two again! Fifty six terms? No. One line. That's it!

5:36Example 2: x^5 in (1+x)^21 + (1+x)^22 + ... + (1+x)^30

Now a harder one. Find the coefficient of x to the power five in one plus x to the power twenty one, plus one plus x to the power twenty two, and so on, up to one plus x to the power thirty. Pause the video and try it. Hint: where does the sum start? Done? Okay, let's check. From each bracket: twenty one C five, plus twenty two C five, and so on, up to thirty C five. Okay? So, like example one, the answer is thirty one C six, right? Oh no, careful! This sum starts at twenty one C five, not at five C five on the edge. The terms from five C five up to twenty C five are missing, so we subtract them. Clear? So use our closed form, with m twenty one, n thirty and r five. The sum of k C five, for k from twenty one to thirty, equals thirty one C six, minus twenty one C six. Leave it in this form. Right? Now check it the G P way. The sum is one plus x to the power thirty one, minus one plus x to the power twenty one, all over x. Because of the x below, we need x to the power six in the numerator. Got it? And there, the coefficient of x to the power six is thirty one C six, minus twenty one C six. Same answer, both ways! Very good.

7:03Quick revision: hockey stick and the G.P. route

So, quick revision. One: for x to the power r, add the k C r from each bracket. Two: the hockey stick: the sum of k C r, for k from m to n, is n plus one C r plus one, minus m C r plus one. Three: the G P route divides by x, so find x to the power r plus one in the numerator. Four: the sum of x to the power k, times one plus x to the power n minus k, is one plus x to the power n plus one, minus x to the power n plus one. Now go and try the practice problems on this page. All the best!