Weighted Sums: ΣrCᵣ and Relatives

Transcript

0:00Introduction

Hello everyone! Okay, a quick challenge. One times ten C one, plus two times ten C two, and so on, up to ten times ten C ten. The terms are ten, ninety, three hundred and sixty, and the middle ones go past a thousand. Can you add all ten in your head? Tough, right? Soon it will take one line. Weighted sums like this come up often in JEE, so let's begin.

0:27Method 1: absorb the weight r

So, C r means n C r, with n a positive integer. First tool: absorption, from earlier in this unit. r times C r equals n times n minus one C r minus one, with n minus one small at the top left, and r minus one at the bottom right. The weight r is absorbed, and n comes out. Okay? Now add from r equals one, below the sigma, to n, on top. The big Greek letter sigma means add them all, and n comes outside it. Here r minus one runs from zero to n minus one. So what do all these n minus one C's add up to? Yes, exactly! Two to the power, n minus one. So the sum of r C r is n times two to the power, n minus one. Write this down. Clear? Now our challenge: n is ten. Ten times two to the power nine; the dot means times. That is five thousand one hundred and twenty. One line! Got it?

1:25Method 2: differentiate the expansion

Now the second method. One plus x, in round brackets, to the power n, is the sum of C r times x to the power r, from r equals zero to n. Differentiate both sides with respect to x. That is the d over d x you see on screen. What does that do to x to the power r? Correct! It brings r down in front as a weight, and the power drops to r minus one. The left side becomes n times one plus x, to the power, n minus one. The constant term at r equals zero is gone, so the sum now starts at r equals one. With me? Now put x equals one. Same answer: n times two to the power, n minus one. Two methods, one answer. Perfect!

2:12Bigger weights: the falling-product trick

Now bigger weights. r times r minus one is a falling product: r, then one less. Absorb twice, and it becomes n times n minus one, times n minus two C r minus two. With me? Add them all: the n minus two C's give two to the power, n minus two. So this sum is n times n minus one, times two to the power, n minus two. Okay? Now r squared. Well, r squared is just r times r, so absorb twice again, right? No, no, careful! After one absorption, the C has r minus one at the bottom, so the next weight must be r minus one. So, the topper's trick: r squared equals r times r minus one, plus r. Clear? Each part absorbs: n times n minus one, times two to the power, n minus two, plus n times two to the power, n minus one. Take two to the power, n minus two, common: n times n plus one, times two to the power, n minus two. Got it? And r cubed? Same trick: r times r minus one times r minus two, plus three times r times r minus one, plus r. We finish it in Example two. With me?

3:23Linear weights: pair the ends

Now a very common type: the weight is a times r, plus b, where a and b are fixed numbers. Call this sum capital S, okay? The shortcut: pair the term with C r and the term with C n minus r. By symmetry, C r equals C n minus r. Their weights are a times r, plus b, and a times n minus r, plus b. What do they add up to? Very good! The r cancels! Every pair gives a times n, plus two b. With me? So add S forwards and backwards: two S equals a times n, plus two b, times two to the power n. The double arrow means so: S equals a times n, plus two b, times two to the power, n minus one. Clear?

4:08Alternating signs: the sum is zero

One last result. With alternating signs, minus one to the power r, the sum of r C r is zero, for n greater than or equal to two. Okay? Why? Absorb the r: we get n times the sum of minus one to the power r, times n minus one C r minus one. That is minus n, times one minus one, to the power, n minus one. It is zero when this power is at least one, so n must be at least two. Got it?

4:36Example 1: C0 + 4C1 + 7C2 + ... + (3n+1)Cn

Now, Example one. Find C zero, plus four C one, plus seven C two, and so on, up to three n plus one, times C n. Pause the video and try it. Done? Okay, let's check. The weights one, four, seven go up by three, so the weight on C r is three r plus one. Okay? Split it. Three r gives three n times two to the power, n minus one. And the one adds one. So the answer is this, plus one, right? No, no, careful! The plus one sits on every C r, so it adds up all the C r's: two to the power n, not one. So the sum is three n times two to the power, n minus one, plus two to the power n. Clear? Write two to the power n as two times two to the power, n minus one, and take that common: three n plus two, times two to the power, n minus one. With me? And pairing agrees, in one line. Here a equals three and b equals one, so a times n, plus two b, is three n plus two. Very good!

5:40Example 2: sum of r^3 nCr, checked for n = 5

Now a harder one. Show that the sum of r cubed C r, from r equals zero to n, equals n squared, times n plus three, times two to the power, n minus three. Then check it for n equals five. Pause the video and try it. Done? Okay, let's check. First, our trick splits r cubed into three falling products. With me? Absorb each piece. We get n times n minus one times n minus two, times two to the power, n minus three; then three n times n minus one, times two to the power, n minus two; then n times two to the power, n minus one. What is common to all three? Yes, n times two to the power, n minus three! Take it outside the square brackets. The middle term picks up a two, so three becomes six, and the last term picks up a four. Okay? Expand: n squared minus three n plus two, plus six n minus six, plus four. That is n squared plus three n, which is n times n plus three. So the sum is n squared, times n plus three, times two to the power, n minus three. Got it? Now n equals five. n squared is twenty five, n plus three is eight, and two to the power two is four. Twenty five times eight times four: eight hundred. Clear? And directly: five, plus eighty, plus two hundred and seventy, plus three hundred and twenty, plus one hundred and twenty five. Eight hundred again! Perfect!

7:16Quick revision: absorb r, falling products, pair the ends

So, quick revision. One: absorb the weight, and the sum of r C r is n times two to the power, n minus one. Two: split r squared or r cubed into falling products, and absorb each one. Three: for weights a times r, plus b, pair the ends, and the sum is a times n, plus two b, times two to the power, n minus one. Now try the practice problems on this page. All the best!