Sums with frac 1r+1: Integration

Transcript

0:00Introduction

Hello everyone! Quick puzzle: add these six fractions: one, plus five over two, plus ten over three, plus ten over four, plus five over five, plus one over six. Can you do it in ten seconds? Hard, right? But look at the tops: one, five, ten, ten, five, one, the row of Pascal's triangle for n equals five. Today one integral adds them all. Sums like this come up often in JEE, so let's begin.

0:31Dividing by r+1? Integrate!

So, we start from 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, plus C two times x squared, 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? Now, in today's sums, each C r is divided by r plus one. r plus one sits in the denominator. So which operation divides by the new power? Integration, exactly! The long S is the integral sign, with limits zero and one, and d x at the end. The integral from zero to one of x to the power r, d x, is one over r plus one. Right? So, integrate the expansion term by term, from zero to x: the integral of one plus t, to the power n, d t. Here t is just a dummy variable. Each C r times t to the power r gives C r times x to the power, r plus one, over r plus one. The capital sigma adds them, r from zero to n. Clear? And the left side integrates directly: one plus x, to the power, n plus one, minus one, all over n plus one. This is today's master formula. Got it?

1:50Put x=1 and x=-1

Now put x equals one. In the master formula's sum, x to the power, r plus one, becomes just one. So on the left we get our sum: C zero, plus C one over two, plus C two over three, and so on, up to C n over n plus one. Okay? And in the master formula, one plus x becomes one plus one, which is two. So the answer is two to the power, n plus one, over n plus one, right? No, no, careful! You forgot the minus one. It comes from the lower limit zero: one plus zero, to the power, n plus one, is one. So the sum is two to the power, n plus one, minus one, all over n plus one. Classic trap! With me? Now put x equals minus one. One plus x is zero, so the right side is zero minus one, over n plus one. On the left, minus one to the power, r plus one, makes the signs alternate. Right? Multiply both sides by minus one. So the sum of minus one to the power r, times C r, over r plus one, is just one over n plus one. Beautiful!

2:57Second method: absorption read backwards

Now a second way, with no calculus: the absorption rule, written for the row n plus one. r plus one, times n plus one C, r plus one, equals n plus one, times C r. Okay? Read it backwards: divide by r plus one, and by n plus one. C r over r plus one equals n plus one C, r plus one, over n plus one. The awkward denominator is now a fixed number. Clear? Now add, from r equals zero to n. The tops run from n plus one C one, up to n plus one C, n plus one. The full row would give two to the power, n plus one. So which coefficient is missing? n plus one C zero, yes! And it equals one. So the tops add up to two to the power, n plus one, minus one. Divide by n plus one: the same answer. Perfect!

3:52Two factors below: (r+1)(r+2)

And if the denominator is r plus one, times r plus two? Apply the same rule twice. C r, over r plus one times r plus two, equals n plus two C, r plus two, over n plus one, times n plus two. Got it? Now the tops are the row of n plus two, with two coefficients missing: n plus two C zero, which is one, and n plus two C one, which is n plus two. Okay? Take both away from two to the power, n plus two. The sum is two to the power, n plus two, minus n, minus three, all over n plus one, times n plus two. Right?

4:29Example 1: sum of 5Cr/(r+1), checked term by term

Now, let's finish our puzzle. Find the sum of five C r over r plus one, for r from zero to five, and check it by adding the terms. Pause the video and try it. Done? Okay, let's check. Here n is five. So two to the power five, minus one, over five, right? Oh no, careful! The power and the bottom are both n plus one, which is six. So it is two to the power six, minus one, over six: sixty three over six, that is, twenty one over two. With me? Now the check. The terms are one, five over two, ten over three, ten over four, one again, and one over six. Over a common denominator, they add up to twenty one over two again! Puzzle solved! Very good!

5:19Example 2: sum of nCr/(r+2) by integration

Now a harder one. Find the sum of C r over r plus two, for r from zero to n. Pause the video and try it. Hint: think about which power of x you need. Done? Okay, let's check. Integrating x to the power r from zero to one gives one over r plus one. But we want one over r plus two. So which power of x must sit next to C r? x to the power, r plus one, exactly! So multiply the expansion by x. x times one plus x to the power n equals C zero x, plus C one x squared, and so on, up to C n times x to the power, n plus one. Okay? Integrate from zero to one, and each term gives C r over r plus two. So our sum is the integral from zero to one of x times one plus x to the power n, d x. Right? To integrate, write x as one plus x, minus one. The integrand becomes: one plus x, to the power, n plus one, minus, one plus x, to the power n, held together in square brackets. Clear? Each piece is our master formula. The first gives two to the power, n plus two, minus one, over n plus two. The second, two to the power, n plus one, minus one, over n plus one. Got it? Over the common denominator, n plus one times n plus two, the numerator is two to the power, n plus one, times the bracket: two times, n plus one, minus, n plus two. And then plus one. That bracket is just n. So the numerator is n times two to the power, n plus one, plus one. With me? So the answer is that numerator, over n plus one, times n plus two. One small multiplication by x, and the same integral did it all. Perfect!

7:13Quick revision: integrate, or read absorption backwards

So, quick revision. One: r plus one below C r? Integrate one plus x to the power n, from zero to x. Two: the sum of C r over r plus one is two to the power, n plus one, minus one, over n plus one, and the alternating sum is one over n plus one. Three: or read absorption backwards. Four: r plus two below? Multiply by x first. Now go and try the practice problems on this page. All the best!