MEI FP1 Study Resources Induction And Series 1-2

MEI Further Pure Mathematics 1 Induction and series Section 1: Proof by induction Study Plan Background You will have already met the idea of proof in mathematics. If you have covered chapter 6 “The language of mathematics” in the AS Pure Mathematics textbook, then you will have come across several different methods of proof, including proof by exhaustion, proof by deduction and proof by contradiction. These are summarised in the Notes and Examples. In this section you will meet another, very powerful, method of proof called proof by induction. You will be using it to prove formulae for the sums of number series. For example, suppose you want to prove that the sum of the first n odd numbers is given by n². You could check this result for different values of n, but this is not a proof as you cannot check all possible values of n. The method of proof by induction allows you to in effect check all values of n by generalising the checking process.

Detailed work plan 1. Read pages 115 – 119. There is an alternative explanation of proof by induction, using a practical example, in the Notes and Examples. 2. There is also a PowerPoint example showing proof by induction for a simple number series. 3. Exercise 5A Attempt at least Questions 1, 3, 4, 5, 10 and 11. If you need more practice, try some of the others as well. 4. Read page 120. This shows some other uses for proof by induction. Example 5.3 is beyond the syllabus, but it is interesting to look at how induction can be used in this kind of proof. 5. There is a PowerPoint version of Example 5.3. 6. Exercise 5B Attempt at least questions 1, 3, 4 and 5. If you have time you may like to attempt the enrichment questions 7, 8 and 9 as well. Note that there is no multiple choice test for this section.

© MEI, 30/04/09

1/1

MEI Further Pure Mathematics 1 Induction and series Section 1: Proof by induction Notes and Examples These notes contain subsections on  Types of proof: a reminder  Proof by induction

Types of proof: a reminder You have probably already met the idea of proof in your study of mathematics. Some basic ideas about proof and the language of proof are covered in chapter 6 of the AS Pure Mathematics textbook, and there is more about proof in C3. As a reminder, here are some of the basic ideas about proof:  To prove something in mathematics, you need to show that it is true for all possible cases. For example, if you wanted to prove that the sum of the first n odd numbers is given by n², you could check as many different values of n as you like, or you could get a computer to check all values of n up to a very large value, but this would still not prove the result. You might feel pretty confident that the result was correct, but you would not have a proof.  To disprove something in mathematics, you only need to find one example for which it is not true. This is called a counterexample. There are several different types of proof which you may have already come across:  Proof by exhaustion: this is when you check all possible cases. You can’t do this if there is an infinite number of cases, as in the example above about the sum of the first n odd numbers. However, you could use proof by exhaustion to prove that 101 is prime, since you could test to see if 101 is divisible by any number less than 101.  Proof by deduction: this is when you use known results to deduce further results. For example, there are several ways to prove Pythagoras’ theorem, using results you already know, such as the area of a triangle.  Proof by contradiction: here you assume that the result is not true, and use this assumption to deduce a result which is impossible, or contradicts the original assumption. This means that the original assumption must be wrong.

© MEI, 30/04/09

1/5

MEI FP1 Series Section 1 Notes and Examples Proof by induction In this section you meet another method of proof, called proof by induction. Proof by induction is a topic that many students find difficult. In fact it is not really all that hard to actually do the questions; the problem is in understanding how and why proof by induction works. Here is a practical example which may help.

The problem: prove that the maximum number of pieces into which you can n2  n  2 cut a pizza with n cuts is given by . 2 First we need to give some thought to the general principles of successful pizza cutting. If you want to get the maximum number of pieces with the minimum number of cuts, the first important thing is not to allow more than two cuts to meet at the same point. For example, three cuts all meeting at the same point would give six pieces, but three cuts which do not meet at the same point give seven pieces.

WRONG 6 pieces

CORRECT 7 pieces

Secondly, you must make sure that each new cut you make crosses each of the previous cuts. The diagrams below show a fourth cut being added.

WRONG 10 pieces

CORRECT 11 pieces

Let’s now see how this works out for the first four cuts.

If you make 1 cut, you cut the pizza into two pieces.

© MEI, 30/04/09

n=1 2 pieces

2/5

MEI FP1 Series Section 1 Notes and Examples

When you make a second cut, you cut both pieces into two, giving two extra pieces, making a total of 4.

n=2 4 pieces

When you make a third cut, you cross each of the two previous cuts. This means that you are cutting through three existing pieces, so you create three extra pieces.

n=3 7 pieces

When you make a fourth cut, you cross each of the three previous cuts. This means that you are cutting through four existing pieces, so you create four extra pieces.

n=4 11 pieces

You can see that the formula number of pieces from n cuts =

n2  n  2 2

works for the cases where n = 1, 2, 3 and 4. In general, suppose that we already have k cuts and we want to add the (k + 1)th cut. We need to make sure that this cut crosses all k of the cuts we have made so far. This means that we are cutting through k + 1 pieces to create k + 1 new pieces. (We start off cutting a single piece into two, then cross a cut line, then cut a second piece in two, then cross a second cut line, etc., until we have crossed to the other side of the pizza, cutting the last piece we encounter in two. That's one more new piece made than the cuts we crossed). Now that we know how the pattern works, we can continue to check that the formula holds for different numbers of cuts without drawing the pizzas, just adding on from previous results. We know that 4 cuts produces 11 pieces. For 5 cuts, the number of pieces  11  5  16 

52  5  2 2

so the formula is correct for n = 5.

© MEI, 30/04/09

3/5

MEI FP1 Series Section 1 Notes and Examples For 6 cuts, the number of pieces  16  6  22 

62  6  2 2

so the formula is correct for n = 6. For 7 cuts, the number of pieces  22  7  29 

72  7  2 2

so the formula is correct for n = 7. Suppose we want to see if the formula is true for n = 100. We could use the formula to work out the number of pieces for n = 99: 992  99  2  4951 . Number of pieces  2 Then we could use this result to find the result for n = 100. 1002  100  2 Number of pieces  4951  100  5051  . 2 So, the formula seems to work for n = 100, but to work this out we have assumed that the formula works for n = 99. All the above calculation tells us is that IF the formula is true for n = 99, THEN it is true for n = 100. We could check n = 99 in the same way, by assuming that the formula works for n = 98, and showing that adding on 99 gives the correct result for n = 99. This now tells us that IF the formula is true for n = 98, THEN it is true for n = 99. We could carry on working backwards like this, until we get down to a result which we already know is true, such as n = 7. Not a very efficient method of proof, and it doesn’t prove that the result is true for ALL values of n. However, we can generalise this process to show that it is true for all values of n. We assume that the formula is correct for the first k cuts. This means that we k2  k  2 already have pieces. We want to show that for k + 1 cuts, the 2 (k  1) 2  (k  1)  2 number of pieces is given by , which can be simplified to 2 k 2  3k  4 . 2 So, after the (k + 1)th cut, which gives us k + 1 additional pieces, the total number of pieces is given by k2  k  2  k 1 2 k 2  k  2  2k  2 k 2  3k  4  Simplifying gives 2 2 which is the result which we were expecting for k + 1 cuts.

© MEI, 30/04/09

4/5

MEI FP1 Series Section 1 Notes and Examples What we have now shown is that IF the formula is true for n = k, THEN it is true for n = k + 1. This is true for ANY value of k. So, in a few lines, we have shown that: IF the formula is true for n = 99, THEN it is true for n = 100 IF the formula is true for n = 42, THEN it is true for n = 43 IF the formula is true for n = 10, THEN it is true for n = 11 IF the formula is true for n = 13927, THEN it is true for n = 13928 etc. Of course this applies to ALL possible values! This means that all we need to do is to check that the result is true for an initial case, say n = 1, and we can then say: Since we know the formula is true for n = 1, then it must be true for n = 2. Now that we know the formula is true for n = 2, then it must be true for n = 3. Now that we know the formula is true for n = 3, then it must be true for n = 4. And so on. We can continue this as far as we like. So the formula is true for all values of n ≥ 1. Look at Example 5.1 in the FP1 textbook to see how a proof by induction should be formally set out. Notice the three essential steps: Step 1 Step 2 Step 3

Prove that the result is true for a starting value, such as n = 1. Prove that if it is true for n = k, then it is true for n = k + 1. Conclude the argument.

Step 3 is just writing down the couple of sentences shown in the text book example. You MUST include this: marks will be given for it. Remember that n = 1 may not always be the starting point! For a challenge, click here and try to find the fallacy in the “proof by induction”.

There is an interactive PowerPoint example of proof by induction using a simple number series. There is also a PowerPoint version of Example 5.3 in the textbook. This is a divisibility proof, which is beyond the syllabus but an interesting extension of this topic.

© MEI, 30/04/09

5/5

MEI Further Pure Mathematics 1 Induction and series Section 1: Proof by induction Crucial points 1. Understand the concept Make sure that you really understand the principle behind proof by induction. The Notes and Examples should help. 2. Always think about what you are aiming for! When you take the assumed result for n = k and add on the (k + 1)th term, you want to rearrange this to get the formula for n = k + 1. It may help to actually write down the result you are looking for (this is done in Example 5.1 in the book). 3. Be careful with algebraic manipulation It is easy to make mistakes. Thinking about the result you are aiming for (see above) often helps, as it may give you a clue about what factors you could take out. 4. Make sure that you write out the proof correctly Remember that there are three steps involved, and you will lose marks if you don’t, for example, write down the conclusion of the argument (Step 3).

© MEI, 30/04/09

1/1

Further Pure Mathematics 1 Induction and series Section 1: Proof by induction Exercise In Questions 1 to 7 prove the given result by induction. 1. The sum of the first n terms of the series (1 × 3) + (2 × 4) + (3 × 5) +……….. n(n + 1)(2n + 7) is 6 2. The sum of the first n terms of the series n 1 1 1 1 is + + + ..... + 1× 3 3 × 5 5 × 7 (2r − 1)(2r + 1) (2n + 1) n

3.

∑r

2

( r + 1) =

r =1

n

4.

∑2

r −1

n( n + 1)( n + 2)(3n + 1) 12

= 2n − 1

r =1

n

5.

r

∑2 r =1

r

= 2−

(n + 2) 2n

6. Given that u(n+1) = 2un + 1 where n is a positive integer and u1 = 5 un = 3 × 2n – 1. ⎛ 3 − 1⎞ n ⎛ 2n + 1 − n ⎞ ⎟⎟ , A = ⎜⎜ ⎟⎟ where n is a positive integer. 7. If A = ⎜⎜ ⎝ 4 − 1⎠ ⎝ 4n 1 − 2n ⎠

Enrichment questions

8. Prove by induction that n3 + 3n2 – 10n is a multiple of 3 for all positive integers n. 9. Prove by induction that 32n – 1 is a multiple of 8 for all positive integers n.

© MEI, 05/01/07

1/1

Further Pure Mathematics 1 Induction and series Section 1: Proof by induction Solutions to Exercise 1. To prove that

n

r (r + 2) = ∑ r

n(n + 1)(2n + 7 ) 6

=1

.

Step 1:

When n = 1, L.H.S. = 1 × 3 = 3 1×2× 9 =3 R.H.S. = 6 So the result is true for n = 1.

Step 2:

Assume

Step 2:

k(k + 1)(2k + 7 )

=1

So if the result is true for n = k, then it is true for n = k + 1. Since it is true for n = 1, then it is true for all positive integers greater than or equal to 1 by induction.

2. To prove that

Step 1:

r (r + 2) = ∑ r

6 k(k + 1)(2k + 7) r (r + 2) = + (k + 1)(k + 3) ∑ 6 r =1 k(k + 1)(2k + 7) + 6(k + 1)(k + 3) = 6 2 (k + 1)(2k + 7 k + +6k + 18) = 6 2 (k + 1)(2k + 13k + 18) = 6 (k + 1)(k + 2)(2k + 9) = 6 (k + 1)((k + 1) + 1 ) ( 2(k + 1) + 7 ) = 6

k +1

Step 3:

k

n

1

n

. = ∑ (2n + 1) r (2r − 1)(2r + 1) =1

1 1 = 1×3 3 1 R.H.S. = 3 So the result is true for n = 1. k 1 k Assume ∑ = (2k + 1) r = 1 (2r − 1)(2r + 1)

When n = 1, L.H.S. =

© MEI, 05/01/07

1/5

Further Pure Mathematics 1 k +1

k

1

1

∑ (2r − 1)(2r + 1) = (2k + 1) + (2k + 1)(2k + 3) r =1

k(2k + 3) + 1 (2k + 1)(2k + 3) 2k 2 + 3k + 1 = (2k + 1)(2k + 3) (2k + 1)(k + 1) = (2k + 1)(2k + 3) k+1 = 2(k + 1) + 1 =

Step 3:

So if the result is true for n = k, then it is true for n = k + 1. Since it is true for n = 1, then it is true for all positive integers greater than or equal to 1 by induction.

3. To prove that

n

∑r r

2

=1

(r + 1) =

n(n + 1)(n + 2)(3n + 1) 12

Step 1:

When n = 1, L.H.S. = 1 2 × 2 = 2 1×2 ×3×4 R.H.S. = =2 12 So the result is true for n = 1.

Step 2:

Assume

∑r r

2

(r + 1) =

k(k + 1)(k + 2)(3k + 1)

12 k(k + 1)(k + 2)(3k + 1) r 2(r + 1) = + (k + 1)2(k + 2) ∑ 12 r =1 k(k + 1)(k + 2)(3k + 1) + 12(k + 1)2(k + 2) = 12 2 (k + 1)(k + 2)(3k + k + 12k + 12) = 12 (k + 1)(k + 2)(3k 2 + 13k + 12) = 12 (k + 1)(k + 2)(k + 3)(3k + 4) = 12 (k + 1)((k + 1) + 1 )((k + 1) + 2 )( 3(k + 1) + 1 ) = 12

k +1

Step 3:

k

.

=1

So if the result is true for n = k, then it is true for n = k + 1. Since it is true for n = 1, then it is true for all positive integers greater than or equal to 1 by induction.

© MEI, 05/01/07

2/5

Further Pure Mathematics 1 4. To prove that

n

2r ∑ r

−1

=1

= 2n − 1 .

Step 1:

When n = 1, L.H.S. = 2 0 = 1 R.H.S. = 2 1 − 1 = 1 So the result is true for n = 1.

Step 3:

Assume k +1

2r ∑ r

−1

=1

k

2r ∑ r

−1

=1

= 2k − 1

= 2 k − 1 + 2( k + 1)− 1 = 2k − 1 + 2k = 2 × 2k − 1 = 2k +1 − 1

Step 3:

So if the result is true for n = k, then it is true for n = k + 1. Since it is true for n = 1, then it is true for all positive integers greater than or equal to 1 by induction.

5. To prove that

Step 1:

n

r

∑ 2r r =1

=2−

(n + 2 ) . 2n

When n = 1, L.H.S. =

1 2

3 1 = 2 2 So the result is true for n = 1.

R.H.S. = 2 −

(k + 2 ) 2k =1 k +1 r (k + 2) k + 1 =2− + k +1 ∑ r 2k 2 r =1 2 2(k + 2) − (k + 1) =2− 2k +1 2k + 4 − k − 1 =2− 2k +1 (k + 1) + 2 =2− 2k +1 k

r

∑ 2r r

=2−

Step 2:

Assume

Step 3:

So if the result is true for n = k, then it is true for n = k + 1. Since it is true for n = 1, then it is true for all positive integers greater than or equal to 1 by induction.

© MEI, 05/01/07

3/5

Further Pure Mathematics 1 6. To prove that for un + 1 = 2un + 1 and u 1 = 5 , where n is a positive integer,

un = 3 × 2 n − 1 . Step 1:

When n = 1, u 1 = 3 × 2 1 − 1 = 5 So the result is true for n = 1.

Step 2:

Assume uk = 3 × 2 k − 1 uk + 1 = 2uk + 1 = 2(3 × 2 k − 1) + 1 = 3 × 2k +1 − 2 + 1 = 3 × 2k +1 − 1

Step 3:

So if the result is true for n = k, then it is true for n = k + 1. Since it is true for n = 1, then it is true for all positive integers greater than or equal to 1 by induction.

−n ⎞ ⎛ 3 −1 ⎞ ⎛ 2n + 1 , An = ⎜ where n is a positive 7. To prove that if A = ⎜ ⎟ 1 − 2n ⎟⎠ ⎝ 4 −1 ⎠ ⎝ 4n integer.

Step 1:

Step 2:

−1 ⎞ ⎛ 3 −1 ⎞ ⎛2 × 1 + 1 = =A When n = 1, A 1 = ⎜ 1 − 2 × 1 ⎟⎠ ⎜⎝ 4 −1 ⎟⎠ ⎝ 4× 1 So the result is true for n = 1. −k ⎞ ⎛ 2k + 1 Assume Ak = ⎜ 1 − 2k ⎟⎠ ⎝ 4k −k ⎞ ⎛ 3 −1 ⎞ ⎛ 2k + 1 Ak + 1 = ⎜ 1 − 2k ⎠⎟ ⎝⎜ 4 −1 ⎠⎟ ⎝ 4k −(2k + 1) + k ⎞ ⎛ 3(2k + 1) − 4k =⎜ ⎟ ⎝ 12k + 4(1 − 2k ) −4k − (1 − 2k )⎠ ⎛ 2k + 3 −k − 1 ⎞ =⎜ ⎟ ⎝ 4k + 4 −2k − 1 ⎠ −(k + 1) ⎞ ⎛ 2(k + 1) + 1 =⎜ 1 − 2(k + 1)⎟⎠ ⎝ 4(k + 1)

Step 3:

So if the result is true for n = k, then it is true for n = k + 1. Since it is true for n = 1, then it is true for all positive integers greater than or equal to 1 by induction.

© MEI, 05/01/07

4/5

Further Pure Mathematics 1 8. To prove that n 3 + 3n 2 − 10n is a multiple of 3. Step 1:

When n = 1, n 3 + 3n 2 − 10n = 1 + 3 − 10 = −6 which is a multiple of 3. So the result is true for n = 1.

Step 2:

Assume f(k ) = k 3 + 3k 2 − 10k is a multiple of 3. f(k + 1) = (k + 1)3 + 3(k + 1)2 − 10(k + 1)

= k 3 + 3k 2 + 3k + 1 + 3k 2 + 6k + 3 − 10k − 10 = k 3 + 6k 2 − k − 6 = ( f(k ) − 3k 2 + 10k ) + 6k 2 − k − 6 = f(k ) + 3k 2 + 9k − 6 = f(k ) + 3(k 2 + 3k − 2) Since 3(k 2 + 3k − 2) is a multiple of 3, then if f(k) is a multiple of 3, then f(k+1) is a multiple of 3. Step 3:

So if the result is true for n = k, then it is true for n = k + 1. Since it is true for n = 1, then it is true for all positive integers greater than or equal to 1 by induction.

9. To prove that 3 2 n − 1 is a multiple of 8. Step 1:

When n = 1, 3 2 n − 1 = 3 2 − 1 = 8 which is a multiple of 8. So the result is true for n = 1.

Step 2:

Assume f(k ) = 3 2 k − 1 is a multiple of 8. f(k + 1) = 3 2( k + 1) − 1 = 32k 32 − 1 = 9 × 32k − 1 = 9 ( f(k ) + 1 ) − 1 = 9f(k ) + 8 So if f(k) is a multiple of 8, then f(k+1) is a multiple of 8.

Step 3:

So if the result is true for n = k, then it is true for n = k + 1. Since it is true for n = 1, then it is true for all positive integers greater than or equal to 1 by induction.

© MEI, 05/01/07

5/5

MEI Further Pure Mathematics 1 Induction and series Section 2: Summing series Study Plan Background In this section you will learn two useful techniques for finding the sum of a finite series. The method of differences involves writing a given sequence as the difference of two (or more) other sequences, so that most of the terms cancel out. The second technique is to use standard results for the series n

r , r 1

n

 r 2 and r 1

n

r

3

.

r 1

Detailed work plan 1. Read pages 122 – 124, working through the examples carefully. 2. Exercise 5C Attempt at least Questions 1, 3, 5, 7, 8 and 9. If you have time it is worth doing Questions 10 and 11 as well, as they, with Question 9, provide an introduction to the next section. If you would like a challenge, try Question 12 as well! 3. Read pages 126 and 127. There are some alternative proofs of the n

standard results for

n

r , r r 1

r 1

n

2

and

r

3

in the Notes and Examples.

r 1

4. Exercise 5D Attempt at least Questions 1-7. If you have time you may like to try Question 8 and the starred questions 9-11. You can also try the Series Activity.

© MEI, 30/04/09

1/1

MEI Further Pure Mathematics 1 Induction and series Section 2: Summing series Notes and Examples These notes contain subsections on  The method of differences  Standard results

The method of differences In the method of differences, a series is expressed as the difference of two (or more) other sequences, so that most of the terms cancel out. Examples 5.4 and 5.5 in the FP1 textbook show how this works. Not all series can be summed in this way. The main difficulty in this method is deciding whether it can be used, and seeing how to express the series so that terms do cancel out. However, you will normally be told how to do this (as in the examples in the book and the exercise questions), so this is not a problem for you!

Using standard results The second technique you need to know involves using the standard results n

for the series

r , r 1

n

form

 (ar

3

n

 r 2 and r 1

n

r

3

. Any series which can be expressed in the

r 1

 br 2  cr  d ) can be summed in this way. Examples 5.6 and 5.7

r 1

show how this is done. The tricky part is expressing the final result neatly – you are normally expected to give your answer in a fully factorised form. Look for common factors which you can take out to start with, and take out any fractions as well. Look carefully at the algebraic manipulation in Examples 5.6 and 5.7. For practice in understanding and using the sigma notation, try the Series Activity. n

There are many methods of proving the standard results for

r , r 1

n

r

2

and

r 1

n

r

3

. In Exercise 5C Questions 9-11 these results are proved using the

r 1

method of differences. The results can also be proved by induction (the result

© MEI, 30/04/09

1/4

MEI FP1 Series Section 2 Notes and Examples n

for  r 2 was proved by induction in Example 5.1, r 1

n

 r and r 1

n

r

3

form

r 1

Questions 2 and 4 in Exercise 5A). Here are some alternative proofs, for interest. n

1.

Proof of

r  r 1

1 2

n(n  1) using the sum of an arithmetic series

If you have covered the work on arithmetic series in C2, you will know that an arithmetic series is a series in which each term differs from the last by a fixed number, the common difference. You will also know that the sum of an arithmetic series with n terms is given by Sn  12 n[2a  (n  1)d ] , where a is the first term and d is the common difference. n

 r  1  2  3  ...  n is an arithmetic series with a = 1 and d = 1. r 1

n

So

r  r 1

1 2

n[2a  (n  1)d ]

 12 n[2  (n  1) 1]  12 n[2  n  1]  12 n(n  1) n

2.

Proof of

r  r 1

1 2

n(n  1) using triangle numbers

The nth triangle number, Tn, is the number of dots in a triangular array having n rows, with 1 dot on the top row, 2 dots on the second and so on, and n dots on the nth row. The sum 1 + 2 + 3 + … + n is therefore Tn.

The triangle for Tn can be put next to the same triangle drawn upside down, to form a rectangle with n rows and n + 1 columns.

The number of dots in this rectangle is given by n(n + 1), so the nth triangle number Tn  12 n(n  1)

© MEI, 30/04/09

2/4

MEI FP1 Series Section 2 Notes and Examples n

3.

Proof of

r r 1

2

 16 n(n  1)(2n  1)

The diagram above shows that r 2  1  2  3  ...  (r  1)  r  (r  1)  ...  3  2  1 n

So

r

2

can be expressed as

r 1

n

r

2

 1  (1  2  1)  (1  2  3  2  1)  (1  2  3  4  3  2  1)  ...  (1  2  ...  r  ...  2  1)

r 1

n

So each of the three diagrams below represents

r

2

.

r 1

... ... ... ... ... ...

1 2 3 4 5 ...

1 2 3 4 4 ...

1 2 3 3 3 ...

1 2 2 2 2 ...

1 1 1 1 1 ...

1 1 2 1 2 3 ... ... ...

1 2 3 4 ...

1 2 3 4 5 ...

1 2 3 4 ...

1 1 1 1 1 ...

1 2 1 3 2 1 ... ... ...

1 2 2 2 2 ...

1 2 3 3 3 ...

1 2 3 4 4 ...

1 2 3 4 5 ...

... ... ... ... ... ...

Now imagine that you can slide the left and right diagrams on to the centre diagram so that the shaded boxes shown below overlap … ... ... ... ... ... ...

1 2 3 4 5 ...

1 2 3 4 4 ...

1 2 3 3 3 ...

1 2 2 2 2 ...

1 1 1 1 1 ...

1 1 2 1 2 3 1 2 3 4 ... ... ... ...

1 2 3 4 5 ...

1 2 3 4 ...

1 1 1 1 1 ...

1 2 1 3 2 1 ... ... ...

1 2 2 2 2 ...

1 2 3 3 3 ...

1 2 3 4 4 ...

1 2 3 4 5 ...

... ... ... ... ... ...

… and add the overlapping numbers: ... ... ... ... ... ...

1 2 3 4 5 ...

1 2 3 4 5 ...

1 2 3 4 5 ...

1 2 3 4 5 ...

1 2 3 4 5 ...

1 2 3 4 5 ...

1 2 3 4 5 ...

1 2 3 4 5 ...

© MEI, 30/04/09

1 2 3 4 5 ...

... ... ... ... ... ...

3/4

MEI FP1 Series Section 2 Notes and Examples n

The grid above has 2n + 1 columns, and so it represents (2n  1) r . r 1

n

The original three grids each represent

r

2

, so this gives

r 1

n

n

3 r  (2n  1) r 2

r 1

r 1

 (2n  1) 12 n(n  1)  12 n(n  1)(2n  1) n

r

2

r 1

 16 n(n  1)(2n  1)

n

4.

Proof of

r r 1

3

 14 n 2 (n  1) 2 geometrically

Since r  r  r 2 , then r³ can be represented as the area of r squares of side r. 3

n

This means that

r

3

is the total area of:

r 1

1 square of side 1 + 2 squares of side 2 + 3 squares of side 3 + …. + n squares of side n. These can be arranged as shown below up to n = 5.

1 2

3

4

5

The even sided squares overlap, but there is a gap (unshaded) of exactly the same size as the overlap, so the overlapping regions can be moved to fill the gaps. This shows that n

 r  1  2  3  ...  n  3

2

r 1

 n   r  r 1 

2

  12 n(n  1) 

2

 14 n 2 (n  1) 2

© MEI, 30/04/09

4/4

Further Pure Mathematics 1 Induction and series Section 2: Summing series Crucial points 1. Make sure you have the correct left-over terms when using the method of differences After most of the terms cancel out, the left over terms may not necessarily be the first and the last, and there may be more than two. You need to write out the first few terms and the last few terms in full to spot the pattern. 2. Check your answer by substituting for n Whether using the method of differences or standard results to find a sum of the first n terms of a series, it is a good idea to substitute n = 1, and perhaps n = 2 as well, to check your result. 3. Factorise where possible When using standard results, there can be quite a lot of algebra involved in simplifying the result. Make sure you take out any common factors first, as this makes the algebra a lot simpler.

© MEI, 05/01/07

1/1

Further Pure Mathematics 1 Induction and series Section 2: Summing series Exercise 1. Show that n

1 1 1 1 and hence find ≡ − + (r + 1)(r + 2)(r + 3) 2(r + 1) (r + 2) 2(r + 3) 1

∑ (r + 1)(r + 2)(r + 3) . r =1

2. Show that r(r + 1)(r + 2) – (r – 1)r(r + 1) ≡ 3r(r + 1) and hence find n

∑ r (r + 1) . r =1

3. Show that r(r + 1)(r + 2)(r + 3) – (r – 1)(r)(r + 1)(r + 2) ≡ 4r(r + 1)(r + 2) and hence find the sum of the first n terms of r(r + 1)(r + 2). 4. Show that

4 1 1 1 ≡ − + and use (2r − 1)(2r + 1)(2r + 3) 2(2r − 1) (2r + 1) 2(2r + 3) n

this to find

1

∑ (2r − 1)(2r + 1)(2r + 3) . r =1

n

5. Using the standard results, find an expression for

∑ (r + 1)(2r + 1) and hence r =1

evaluate 2 × 3 + 3 × 5 + 4 × 7 + ………. + 21 × 41. 6. Using the standard results, find the sum of the first n terms of (r + 4)3. 7. Write down an expression for the rth term of the series 1 × 2 + 3 × 4 + 5 × 6 + ……. and hence find the sum of the first n terms. 8. Write down an expression for the rth term of the series 22 + 52 + 82 + 112 + …… and hence find the sum of the first n terms. 9. If S = x + 2x2 + 3x3 + ….. + nxn where x ≠ 1, find S. (Hint: consider S – xS) n

10. Using the standard result for

∑ r 3 , show that 1

2n

∑ r3 =

r = n +1

© MEI, 08/01/07

n 2 (3n + 1)(5n + 3) . 4

1/1

Further Pure Mathematics 1 Induction and series Section 2: Summing series Solutions to Exercise 1.

1 1 1 (r + 2)(r + 3) − 2(r + 1)(r + 3) + (r + 1)(r + 2) − + = 2(r + 1) (r + 2) 2(r + 3) 2(r + 1)(r + 2)(r + 3) 2 r + 5 r + 6 − 2r 2 − 8r − 6 + r 2 + 3r + 2 = 2(r + 1)(r + 2)(r + 3) 2 = 2(r + 1)(r + 2)(r + 3) 1 = (r + 1)(r + 2)(r + 3) n 1 1 2 1 ⎛ ⎞ = − + ∑ ∑ ⎜ ⎟ 2(r + 2) 2(r + 3) ⎠ r = 1 (r + 1)(r + 2)(r + 3) r = 1 ⎝ 2(r + 1) n

⎛1 2 1 ⎞ ⎛1 2 1 ⎞ ⎛ 1 2 1 ⎞ + + =⎜ − + ⎟+⎜ − ⎟+ ⎟+⎜ − 10 ⎠ ⎝ 8 10 12 ⎠ ⎝4 6 8 ⎠ ⎝6 8 ⎛ 1 ⎞ 2 1 − + ... + ⎜ ⎟ 2(n + 1) 2(n + 2) ⎠ ⎝ 2n ⎛ ⎞ 1 2 1 − + +⎜ ⎟ ⎝ 2(n + 1) 2(n + 2) 2(n + 3) ⎠ 1 2 1 1 2 1 = − + + − + 4 6 6 2(n + 2) 2(n + 2) 2(n + 3) 1 1 1 = − + 12 2(n + 2) 2(n + 3) 2. r (r + 1)(r + 2) − (r − 1)r (r + 1) = r (r + 1)(r + 2 − (r − 1))

= r (r + 1)(r + 2 − r + 1) = 3r (r + 1) n

n

=1

=1

(r (r + 1)(r + 2) − (r − 1)r (r + 1)) ∑ 3r (r + 1) = ∑ r r = (1 × 2 × 3) − (0 × 1 × 2) + (2 × 3 × 4) − (1 × 2 × 3) + ... + n(n + 1)(n + 2) − (n − 1)n(n + 1) n

r (r + 1) = ∑ r =1

= n(n + 1)(n + 2) 1 3

n(n + 1)(n + 2 )

© MEI, 08/01/07

1/4

Further Pure Mathematics 1 3. r (r + 1)(r + 2)(r + 3) − (r − 1)r (r + 1)(r + 2) = r (r + 1)(r + 2)(r + 3 − (r − 1)) = 4r (r + 1)(r + 2) n

n

r =1

r =1

∑ 4r (r + 1)(r + 2) = ∑ (r (r + 1)(r + 2)(r + 3) − (r − 1)r (r + 1)(r + 2)) = (1 × 2 × 3 × 4) − (0 × 1 × 2 × 3) + (2 × 3 × 4 × 5 ) − (1 × 2 × 3 × 4) +... + n(n + 1)(n + 2)(n + 3) − (n − 1)n(n + 1)(n + 2) = n(n + 1)(n + 2)(n + 3)

n

r (r + 1)(r + 2) = ∑ r =1

4.

1 4

n(n + 1)(n + 2)(n + 3)

1 1 (2r + 1)(2r + 3) − 2(2r − 1)(2r + 3) + (2r − 1)(2r + 1) + = 2(2r − 1) (2r + 1) 2(2r + 3) 2(2r − 1)(2r + 1)(2r + 3) 2 4r + 8r + 3 − 8r 2 − 8r + 6 + 4r 2 − 1 = 2(2r − 1)(2r + 1)(2r + 3) 8 = 2(2r − 1)(2r + 1)(2r + 3) 4 = (2r − 1)(2r + 1)(2r + 3) 1

−

n 4 1 2 1 ⎛ ⎞ = − + ∑ ∑ ⎜ 2(2r + 1) 2(2r + 3) ⎟⎠ r = 1 (2r − 1)(2r + 1)(2r + 3) r = 1 ⎝ 2(2r − 1) n

1 ⎞ ⎛ 1 2 1 ⎞ 1 ⎞ ⎛1 2 ⎛1 2 + − + +⎜ − =⎜ − + ⎟ ⎟+⎜ ⎟ ⎝ 2 6 10 ⎠ ⎝ 6 10 14 ⎠ ⎝ 10 14 18 ⎠ ⎛ ⎞ 2 1 1 − + ... + ⎜ + ⎟ ⎝ 2(2n − 3) 2(2n − 1) 2(2n + 1) ⎠ ⎛ ⎞ 2 1 1 − + +⎜ ⎟ ⎝ 2(2n − 1) 2(2n + 1) 2(2n + 3) ⎠ 1 2 1 1 2 1 = − + + − + 2 6 6 2(2n + 1) 2(2n + 1) 2(2n + 3) 1 1 1 = − + 3 2(2n + 1) 2(2n + 3) n 1 1⎛1 1 1 ⎞ = ⎜ − + ∑ ⎟ 4 ⎝ 3 2(2n + 1) 2(2n + 3) ⎠ r = 1 (2r − 1)(2r + 1)(2r + 3) 1 1 1 = − + 12 8(2n + 1) 8(2n + 3)

© MEI, 08/01/07

2/4

Further Pure Mathematics 1 5.

n

n

r =1

r =1 n

∑(r + 1)(2r + 1) = ∑(2r 2 + 3r + 1) n

n

r =1

r =1

= 2 ∑ r 2 + 3∑ r + ∑ 1 r =1

= 2 × 61 n(n + 1)(2n + 1) + 3 × 21 n(n + 1) + n = 61 n ( 4n 2 + 6n + 2 + 9n + 9 + 6 ) = 61 n(4n 2 + 15 n + 17)

Putting n = 20: 2 × 3 + 3 × 5 + 4 × 7 + ………. + 21 × 41 = 61 × 20(4 × 20 2 + 15 × 20 + 17) = 6390

6.

n

(r + 4) ∑ r

3

=

=1

n +4

r ∑ r

r =4

− ∑r 3

3

=1

r =1

= (n + 4)2(n + 5 )2 − 41 × 42 × 5 2 1 4

((n 2 + 9n + 20)2 − 400 ) = 41 (n 4 + 18n 3 + 121n 2 + 360n + 400 − 400 ) =

1 4

= 41 n(n 3 + 18n 2 + 121n + 360)

7. The rth term is 2r(2r – 1). n

n

n

r =1

r =1

∑ ( 2r (2r − 1)) = 4∑ r 2 − 2 ∑ r r =1

= 4 × n(n + 1)(2n + 1) − 2 × 21 n(n + 1) 1 6

= 31 n(n + 1)( 4n + 2 − 3 ) = 31 n(n + 1)(4n − 1) 8. The rth term is (3r – 1)² n

n

(3r − 1) = ∑( 9r ∑ r r =1

2

=1

n

2

− 6r + 1) n

n

r =1

r =1

= 9∑ r 2 − 6∑ r + ∑ 1 r =1

= 9 × n(n + 1)(2n + 1) − 6 × 21 n(n + 1) + n 1 6

= 21 n ( 6n 2 + 9n + 3 − 6n − 6 + 2 ) = 21 n(6n 2 + 3n − 1)

© MEI, 08/01/07

3/4

Further Pure Mathematics 1 9. S = x + 2 x 2 + 3 x 3 + ..... + nx n

xS = x 2 + 2 x 3 + 3 x 4 + ..... + nx n + 1 S − xS = x + x 2 + x 3 + ... + x n − nx n + 1 x + x 2 + x 3 + ... + x n is a geometric progression with common ratio x x(1 − x n ) so x + x 2 + x 3 + ... + x n = 1−x n x(1 − x ) S − xS = − nx n + 1 1−x x − x n + 1 − nx n + 1 + nx n +2 S(1 − x ) = 1−x x − (n + 1)x n + 1 + nx n + 2 S= (1 − x )2

10.

n

∑r

3

1 2n

∑

r =n + 1

= 41 n 2(n + 1)2 2n

n

r 3 = ∑r 3 − ∑r 3 r =1

r =1

= (2n ) (2n + 1)2 − 41 n 2(n + 1)2 1 4

2

= 41 n 2 ( 4(2n + 1)2 − (n + 1)2 ) = 41 n 2(16n 2 + 16n + 4 − n 2 − 2n − 1) = 41 n 2(15 n 2 + 14n + 3) = 41 n 2(3n + 1)(5 n + 3)

© MEI, 08/01/07

4/4

Series Activity Cut out the cards and match up expressions which are equivalent. Try to justify your decisions algebraically. 9

∑ ( r + 1)( r + 2 ) r=0

10

∑ r (r

− 1)

r =1

11

∑ (r r=2 9

∑

2

− r)

(r 2 + r )

r=0

11

∑ ( r − 1) ( r − 2 ) r=2

30

∑ (r

− 20 )( r − 19 )

r = 21

10

∑ r ( r + 1) r =1

30

∑ ( r − 20 )( r − 21 )

r = 21

© Susan Wall, MEI 2005

1/1

Further Pure Mathematics 1 Induction and series Section 2: Summing series Multiple Choice Test 1) The expression

1 can be written as (2r − 1)(2r + 1)

1 1 − (2r − 1) (2r + 1) 1 1 (c) + 2(2r − 1) 2(2r + 1) (e) I don’t know

1 1 − 2(2r − 1) 2(2r + 1) 1 1 (d) + (2r − 1) (2r + 1)

(a)

(b)

n

2) Using the result from Question 1, the sum

1

∑ (2r − 1)(2r + 1)

is given by

r =1

2n 2n + 1 n +1 (c) 2n + 1 (e) I don’t know

(a)

n 2(2n + 1) n (d) 2n + 1

(b)

8 1 2 1 , = − + (2r − 1)(2r + 1)(2r + 3) 2r − 1 2r + 1 2r + 3 8 8 8 8 the sum + + + ... + 1× 3 × 5 3 × 5 × 7 5 × 7 × 9 (2n − 1)(2n + 1)(2n + 3) is given by

3) Using the result

2 1 1 − + 3 2n + 1 2n + 3 1 2 1 (c) − + 3 2n + 1 2n + 3 (e) I don’t know

(a)

1 2n + 3 1 (d) 1 − 2n + 3

(b) 1 +

© MEI, 05/01/07

1/3

Further Pure Mathematics 1 4) Using the result 3r 2 + 3r + 1 = (r + 1)3 − r 3 , the sum 1 1 + 8 (n + 1)3 1 (c) 1 − (n + 1)3 (e) I don’t know

(b) 1 −

(a)

(d)

n

In Question 5-7, S1 denotes

∑ r , S2 denotes r =1

∑ ( 4r n

5) The sum

r =1

2

3r 2 + 3r + 1 is given by ∑ 3 3 r =1 ( r + 1) r n

1 n3

7 1 − 8 ( n + 1)3

n

∑ r 2 and S3 denotes r =1

n

∑r

3

.

r =1

(r + 1) + 2 ) is equal to

(a) 4S3 + 4S2 + 2n (c) 4S3 + S2 + 2n (e) I don’t know

(b) 4S3 + 4S2 + 2 (d) 4S2(S1 + 1) + 2

n

6) The sum

∑ ( (r + 1)(r − 1) + 1) is equal to r =1

(a) (S1 + 1)(S1 – 1) + 1 (c) (S1 + 1)(S1 – 1) + n (e) I don’t know

(b) S2 (d) S1² – n² + n

n

7) The sum

∑ (3r − 1)(2r + 1) is equal to r =1

(a) 6S2 + S1 – 1 (c) 6S2 – S1 – n (e) I don’t know

(b) 6S2 + S1 – n (d) (3S1 – 1)(2S1 + 2)

n

8) The sum

∑ (2r + 1)(2r − 1)

=

r =1

(a) 13 n(4n2 + 6n − 1) (c) 23 n(n + 1)(2n + 1) − 1

(b) n 2 (n + 1) 2 − 1 (d) 13 n(2n 2 + 3n − 2)

(e) I don’t know

© MEI, 05/01/07

2/3

Further Pure Mathematics 1 n

9) The sum

∑ r (r + 2)(r − 2) = r =1

(a) 14 n(n + 1)(n 2 + n − 8)

(b) 14 n(n + 1)(n − 1)(n + 2)

(c) 14 n2 (n + 1) 2 − 4n (e) I don’t know

(d) 14 n 2 (n + 1) 2 − 4

10) The sum (1× 3) + (2 × 5) + (3 × 7) + ... + n(2n + 1) = (a) 16 n(n + 1)(n + 2) (c) 13 n(n + 1)(n + 2) (e) I don’t know

(b) 12 n(n + 1)(n 2 + n + 1) (d) 16 n(n + 1)(4n + 5)

© MEI, 05/01/07

3/3

Further Pure Mathematics 1 Induction and series Chapter assessment n

1. Find

∑ (6r − 1)(4r − 1) , giving your answer in its simplest form.

[5]

r =1

2. Prove by induction that n

∑ (3r r =1

3. (i) Show that

− 2r ) = 12 n(n + 1)(2n − 1)

[7]

1 1 1 . − = r + 1 r + 2 (r + 1)(r + 2)

[1]

2

(ii) Hence find the sum of the series

n

1

∑ (r + 1)(r + 2) .

[4]

r =1

4. A sequence of real numbers u1 , u2 , u3 ,... is defined by 3un u1 = −1 and un +1 = for n ≥ 1. 5un + 6 3 Prove by induction that un = n . 2 −5 5. Find the sum of the series 1× 9 + 2 × 33 + 3 × 73 + ... + n(8n 2 + 1) , giving your answer in a fully factorised form. 1 5 4 3r − 2 6. (i) Show that − + . − = r r + 1 r + 2 r (r + 1)(r + 2) (ii) Hence find the sum of the first n terms of the series 1 4 7 + + + ... . 1× 2 × 3 2 × 3 × 4 3 × 4 × 5

n

7. Prove by induction that

∑ r3

r

r =1

= 34 + 14 (2n − 1)3n +1 .

© MEI, 08/01/07

[7]

[5]

[2]

[6]

[8]

1/2

Further Pure Mathematics 1 1 1 1 . − = 6(3r − 1) 6(3r + 5) (3r − 1)(3r + 5) (ii) Hence find the sum of the series 1 1 1 1 . + + + ... + 2 × 8 5 ×11 8 × 14 (3n − 1)(3n + 5)

8. (i) Show that

9. Prove by induction that

2r ( r 2 − 2) 2n +1 1 = − ∑ 2 2 2 ( n + 2) 2 r =1 ( r + 1) ( r + 2) n

[2]

[5]

[8] Total 60 marks

© MEI, 08/01/07

2/2