1. 1 + 2 + 3 + ··· + n = n(n + 1)(2n + 1) 6 Proof: For n = 1 the PDF

In Exercises 1-15 use mathematical induction to establish the formula for n ? 1. 1. 1. 2. + 2. 2. + 3. 2. + ··· + n. 2. = n(n + 1)(2n + 1).

MATHEMATICAL INDUCTION SEQUENCES and SERIES

?n is always divisible by 30. Use the fact that n. 5. ?n = (n?1)n(n+1)(n. 2. +1) to prove that it is divisible by 2 and 3 as well as 5.

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1 · 3+2 · 4+3 · 5 + ··· + n(n + 2) = n(n + 1)(2n + 7). 6 is valid. Proof: (formal style; it is good to do a few proofs this way) We will use the Principle

Proof by Induction - University of Plymouth

12-Feb-2006 Example 3: for n a natural number prove that: 1) if n ? 2 then n3 ? n is always divisible by 3

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* Prove that. 12 + 22 + ··· + n2 = n(n + 1)(2n + 1). 6 for all positive integers n. Exercise 1.2. * Show that. 1 ?. 1. 2. +. 1. 3. ?

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8.1.3 Some important observations. 1. The total number of terms in the binomial expansion of (a + b)n is n + 1 i.e. one more than the exponent n. 2.

CS240 Solutions to Induction Problems Fall 2009 1. Let P(n) be the

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(a ? 1)(b ? 1)(c ? 1) = abc. ?=(a ? 1)2 ? 4a(a ? 1) = (1 ? a)(1 + 3a). ... (i) f(2n ? t(m)) ? 3(n?1)/2 if 2n + m is divisible by 3;.

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principle of induction P(n) is true for all positive integers n > 1. herefore each term in the product (1+2+22)(1 +3 +3² +3³)(1 +.

Polynomials II - Divisibility and Irreducibles

So now for instance x2 ?2 is not divisible by x? ? 2 over Q since x? ? 2 doesn’t exist over Q Problem 9 Show that x2?2 is irreducible over Q (Hint: Suppose it were reducible What would the degrees of the factors have to be and what does that mean?) Problem 10 Show that x3 ?2 is irreducible over Q (Hint: This is similar

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1+n2 = 0 and (ii) the sequence of terms 1+n2 are decreasing To see (i) notice that we can divide numerator and denominator by n2 to get lim n?? 1 n2 ·n 1 n2 (1+n 2) = lim n?? 1 n 1 n2 +1 = 0 To see (ii) let f(x) = x 1+x2 Then f0(x) = (1+x2)·1?x·2x (1+x2)2 = 1?x2 (1+x2)2 ? 0 for x ? 1 Therefore f is a decreasing

3 Mathematical Induction 31 First Principle of

3 MATHEMATICAL INDUCTION 89 Which shows 5(n+ 1) + 5 (n+ 1)2 By the principle of mathematical induction it follows that 5n+ 5 n2 for all integers n 6 Discussion In Example 3 4 1 the predicate P(n) is 5n+5 n2 and the universe of discourse

How to prove n is true for all integers n 1?

Mathematical induction can be used to prove that a statement about n is true for all integers n ? 1. We have to complete three steps. In the basis step, verify the statement for n = 1. In the inductive hypothesis, assume that the statement holds when n = k for some integer k ? 1.

Does f(n) = n 1+n2 converge?

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Is p divisible by Q over C?

Definition 1Let p and q be polynomials in the complex numbers C. We say that p is divisble by q over C if there exists a polynomial r such that p(z) = q(z)r(z). Problem 1 By long division (which we learned how to do last week), answer the following: • Is x3?2x2+ x?2 divisible by x?2 over C?

Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despic In Exercises 1-15 use mathematical induction to establish the formula forn1.

1.12+ 22+ 32++n2=n(n+ 1)(2n+ 1)6

Proof:

Forn= 1, the statement reduces to 12=1236

and is obviously true.

Assuming the statement is true forn=k:

2+ 22+ 32++k2=k(k+ 1)(2k+ 1)6

;(1) we will prove that the statement must be true forn=k+ 1: 1

2+ 22+ 32++ (k+ 1)2=(k+ 1)(k+ 2)(2k+ 3)6

:(2)

The left-hand side of (2) can be written as

2+ 22+ 32++k2+ (k+ 1)2:

In view of (1), this simplies to:

12+ 22+ 32++k2+ (k+ 1)2=k(k+ 1)(2k+ 1)6

+ (k+ 1)2 k(k+ 1)(2k+ 1) + 6(k+ 1)26 (k+ 1)[k(2k+ 1) + 6(k+ 1)]6 (k+ 1)(2k2+ 7k+ 6)6 (k+ 1)(k+ 2)(2k+ 3)6 Thus the left-hand side of (2) is equal to the right-hand side of (2). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integern.

2.3 + 32+ 33++ 3n=3n+132

Proof:

Forn= 1, the statement reduces to 3 =3232

and is obviously true.

Assuming the statement is true forn=k:

3 + 3

2+ 33++ 3k=3k+132

;(3) we will prove that the statement must be true forn=k+ 1: 3 + 3

2+ 33++ 3k+1=3k+232

:(4) Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despic

The left-hand side of (4) can be written as

3 + 3

2+ 33++ 3k+ 3k+1:

In view of (3), this simplies to:

3 + 32+ 33++ 3k+ 3k+1=3k+132

+ 3k+1

3k+13 + 23k+12

33k+132

3k+232

Thus the left-hand side of (4) is equal to the right-hand side of (4). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integern.

3.13+ 23+ 33++n3=n2(n+ 1)24

Proof:

Forn= 1, the statement reduces to 13=12224

and is obviously true.

Assuming the statement is true forn=k:

3+ 23+ 33++k3=k2(k+ 1)24

;(5) we will prove that the statement must be true forn=k+ 1: 1

3+ 23+ 33++ (k+ 1)3=(k+ 1)2(k+ 2)24

:(6)

The left-hand side of (6) can be written as

3+ 23+ 33++k3+ (k+ 1)3:

In view of (5), this simplies to:

13+ 23+ 33++k3+ (k+ 1)3=k2(k+ 1)24

+ (k+ 1)3 k2(k+ 1)2+ 4(k+ 1)34 (k+ 1)2[k2+ 4(k+ 1)]4 (k+ 1)2(k+ 2)24 Thus the left-hand side of (6) is equal to the right-hand side of (6). This proves the inductive step. Therefore, by the principle of mathematical Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despic induction, the given statement is true for every positive integern.

4.1 + 3 + 6 + 10 ++n(n+ 1)2

=n(n+ 1)(n+ 2)6

Proof:

Forn= 1, the statement reduces to 1 =1236

and is obviously true.

Assuming the statement is true forn=k:

1 + 3 + 6 + 10 ++k(k+ 1)2

=k(k+ 1)(k+ 2)6 ;(7) we will prove that the statement must be true forn=k+ 1:

1 + 3 + 6 + 10 ++(k+ 1)(k+ 2)2

=(k+ 1)(k+ 2)(k+ 3)6 :(8)

The left-hand side of (8) can be written as

1 + 3 + 6 + 10 ++k(k+ 1)2

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

In view of (7), this simplies to:

1 + 3 + 6 + 10 ++k(k+ 1)2

+(k+ 1)(k+ 2)2 k(k+ 1)(k+ 2)6 +(k+ 1)(k+ 2)2 k(k+ 1)(k+ 2) + 3(k+ 1)(k+ 2)6 (k+ 1)(k+ 2)(k+ 3)6 Thus the left-hand side of (8) is equal to the right-hand side of (8). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integern.

5.1 + 4 + 7 ++ (3n2) =n(3n1)2

Proof:

Forn= 1, the statement reduces to 1 =122

and is obviously true.

Assuming the statement is true forn=k:

1 + 4 + 7 ++ (3k2) =k(3k1)2

;(9) we will prove that the statement must be true forn=k+ 1:

1 + 4 + 7 ++ [3(k+ 1)2] =(k+ 1)[3(k+ 1)1]2

:(10) Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despic

The left-hand side of (10) can be written as

1 + 4 + 7 ++ (3k2) + [3(k+ 1)2]:

In view of (9), this simplies to:

[1 + 4 + 7 ++ (3k2)] + (3k+ 1) =k(3k1)2 + (3k+ 1) k(3k1) + 2(3k+ 1)2

3k2+ 5k+ 22

(k+ 1)(3k+ 2)2 The last expression is obviously equal to the right-hand side of (10). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integern.

6.12+ 32+ 52++ (2n1)2=n(2n1)(2n+ 1)3

Proof:

Forn= 1, the statement reduces to 12=1333

and is obviously true.

Assuming the statement is true forn=k:

2+ 32+ 52++ (2k1)2=k(2k1)(2k+ 1)3

;(11) we will prove that the statement must be true forn=k+ 1: 1

2+ 32+ 52++ [2(k+ 1)1]2=(k+ 1)[2(k+ 1)1][2(k+ 1) + 1]3

(12)

The left-hand side of (12) can be written as

2+ 32+ 52++ (2k1)2+ [2(k+ 1)1]2:

In view of (11), this simplies to:

Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despic

12+ 32+ 52++ (2k1)2+ (2k+ 1)2

k(2k1)(2k+ 1)3 + (2k+ 1)2 k(2k1)(2k+ 1) + 3(2k+ 1)23 (2k+ 1)[k(2k1) + 3(2k+ 1)]3 (2k+ 1)(2k2+ 5k+ 3)3 (2k+ 1)(k+ 1)(2k+ 3)3 (k+ 1)[2(k+ 1)1][2(k+ 1) + 1]3 Thus the left-hand side of (12) is equal to the right-hand side of (12). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integern.

7.1 + 5 + 9 + 13 ++ (4n3) = 2n2n

Proof:

Forn= 1, the statement reduces to 1 = 2121 and is obviously true.

Assuming the statement is true forn=k:

1 + 5 + 9 + 13 ++ (4k3) = 2k2k;(13)

we will prove that the statement must be true forn=k+ 1:

1 + 5 + 9 + 13 ++ [4(k+ 1)3] = 2(k+ 1)2(k+ 1):(14)

The left-hand side of (14) can be written as

1 + 5 + 9 + 13 ++ (4k3) + [4(k+ 1)3]:

In view of (13), this simplies to:

[1 + 5 + 9 + 13+ (4k3)] + (4k+ 1) = (2k2k) + (4k+ 1) = 2k2+ 3k+ 1 = (k+ 1)(2k+ 1) = (k+ 1)[2(k+ 1)1] = 2(k+ 1)2(k+ 1): Thus the left-hand side of (14) is equal to the right-hand side of (14). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integern. Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despic

8.2 + 23+ 25++ 22n1=2(22n1)3

Proof:

Forn= 1, the statement reduces to 2 =2(221)3

and is obviously true.

Assuming the statement is true forn=k:

2 + 2

3+ 25++ 22k1=2(22k1)3

;(15) we will prove that the statement must be true forn=k+ 1: 2 + 2

3+ 25++ 22(k+1)1=2(22(k+1)1)3

:(16)

The left-hand side of (16) can be written as

2 + 2

3+ 25++ 22k1+ 22(k+1)1:

In view of (15), this simplies to:

2 + 23+ 25++ 22k1+ 22k+1=2(22k1)3

+ 22k+1

2(22k1) + 322k+13

22k+12 + 322k+13

422k+123

2(222k+11)3

2(22k+21)3

The last expression is obviously equal to the right-hand side of (16). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integern. 9.

1123+1234+1345++1n(n+ 1)(n+ 2)=n(n+ 3)4(n+ 1)(n+ 2)

Proof:

Forn= 1, the statement reduces to1123=14423and is obviously true.

Assuming the statement is true forn=k:

1123+1234+1345++1k(k+ 1)(k+ 2)=k(k+ 3)4(k+ 1)(k+ 2);

(17) Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despic we will prove that the statement must be true forn=k+ 1:

1123+1234+1345++1(k+ 1)(k+ 2)(k+ 3)=(k+ 1)(k+ 4)4(k+ 2)(k+ 3):

(18)

The left-hand side of (18) can be written as

1123+1234+1345++1k(k+ 1)(k+ 2)+1(k+ 1)(k+ 2)(k+ 3):

In view of (17), this simplies to:

1123+1234+1345++1k(k+ 1)(k+ 2)

+1(k+ 1)(k+ 2)(k+ 3) k(k+ 3)4(k+ 1)(k+ 2)+1(k+ 1)(k+ 2)(k+ 3) k(k+ 3)2+ 44(k+ 1)(k+ 2)(k+ 3) k(k2+ 6k+ 9) + 44(k+ 1)(k+ 2)(k+ 3) k3+ 6k2+ 9k+ 44(k+ 1)(k+ 2)(k+ 3) (k+ 1)(k2+ 5k+ 4)4(k+ 1)(k+ 2)(k+ 3) k2+ 5k+ 44(k+ 2)(k+ 3) (k+ 1)(k+ 4)4(k+ 2)(k+ 3): Thus the left-hand side of (18) is equal to the right-hand side of (18). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integern.

10.43+ 83+ 123++ (4n)3= 16n2(n+ 1)2

Proof:

Forn= 1, the statement reduces to 43= 161222and is obviously true.

Assuming the statement is true forn=k:

3+ 83+ 123++ (4k)3= 16k2(k+ 1)2;(19)

we will prove that the statement must be true forn=k+ 1: 4

3+ 83+ 123++ [4(k+ 1)]3= 16(k+ 1)2(k+ 2)2:(20)

The left-hand side of (20) can be written as

3+ 83+ 123++ (4k)3+ [4(k+ 1)]3:

Solutions to Exercises on Mathematical Induction Math 1210, Instructor: M. Despic

In view of (19), this simplies to:

43+ 83+ 123++ (4k)3+ 43(k+ 1)3= 16k2(k+ 1)2+ 64(k+ 1)3

= 16(k+ 1)2[k2+ 4(k+ 1)] = 16(k+ 1)2(k+ 2)2: Thus the left-hand side of (20) is equal to the right-hand side of (20). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integern. 11. 15 2+15 4++15

2n=124

1125
n

Proof:

Forn= 1, the statement reduces to15

2=124 1125
. This is equivalent to 125
=124 2425
and is obviously true.

Assuming the statement is true forn=k:

15 2+15 4++15

2k=124

1125
k ;(21) we will prove that the statement must be true forn=k+ 1: 15 2+15 4++15

2(k+1)=124

1125
k+1 :(22)

The left-hand side of (22) can be written as

15 2+15 4++15 2k+15

2(k+1):

In view of (21), this simplies to:

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[PDF] 1. 1 + 2 + 3 + ··· + n = n(n + 1)(2n + 1) 6 Proof: For n = 1 the

How to prove n is true for all integers n 1?

Does f(n) = n 1+n2 converge?

Is p divisible by Q over C?

1.12+ 22+ 32++n2=n(n+ 1)(2n+ 1)6

Proof:

Forn= 1, the statement reduces to 12=1236

Assuming the statement is true forn=k:

2+ 22+ 32++k2=k(k+ 1)(2k+ 1)6

2+ 22+ 32++ (k+ 1)2=(k+ 1)(k+ 2)(2k+ 3)6

The left-hand side of (2) can be written as

2+ 22+ 32++k2+ (k+ 1)2:

In view of (1), this simplies to:

12+ 22+ 32++k2+ (k+ 1)2=k(k+ 1)(2k+ 1)6

2.3 + 32+ 33++ 3n=3n+132

Proof:

Forn= 1, the statement reduces to 3 =3232

Assuming the statement is true forn=k:

2+ 33++ 3k=3k+132

2+ 33++ 3k+1=3k+232

The left-hand side of (4) can be written as

2+ 33++ 3k+ 3k+1:

In view of (3), this simplies to:

3 + 32+ 33++ 3k+ 3k+1=3k+132

3k+13 + 23k+12

33k+132

3k+232

3.13+ 23+ 33++n3=n2(n+ 1)24

Proof:

Forn= 1, the statement reduces to 13=12224

Assuming the statement is true forn=k:

3+ 23+ 33++k3=k2(k+ 1)24

3+ 23+ 33++ (k+ 1)3=(k+ 1)2(k+ 2)24

The left-hand side of (6) can be written as

3+ 23+ 33++k3+ (k+ 1)3:

In view of (5), this simplies to:

13+ 23+ 33++k3+ (k+ 1)3=k2(k+ 1)24

4.1 + 3 + 6 + 10 ++n(n+ 1)2

Proof:

Forn= 1, the statement reduces to 1 =1236

Assuming the statement is true forn=k:

1 + 3 + 6 + 10 ++k(k+ 1)2

1 + 3 + 6 + 10 ++(k+ 1)(k+ 2)2

The left-hand side of (8) can be written as

1 + 3 + 6 + 10 ++k(k+ 1)2

In view of (7), this simplies to:

1 + 3 + 6 + 10 ++k(k+ 1)2

5.1 + 4 + 7 ++ (3n2) =n(3n1)2

Proof:

Forn= 1, the statement reduces to 1 =122

Assuming the statement is true forn=k:

1 + 4 + 7 ++ (3k2) =k(3k1)2

1 + 4 + 7 ++ [3(k+ 1)2] =(k+ 1)[3(k+ 1)1]2

The left-hand side of (10) can be written as

1 + 4 + 7 ++ (3k2) + [3(k+ 1)2]:

In view of (9), this simplies to:

3k2+ 5k+ 22

6.12+ 32+ 52++ (2n1)2=n(2n1)(2n+ 1)3

Proof:

Forn= 1, the statement reduces to 12=1333

Assuming the statement is true forn=k:

2+ 32+ 52++ (2k1)2=k(2k1)(2k+ 1)3

2+ 32+ 52++ [2(k+ 1)1]2=(k+ 1)[2(k+ 1)1][2(k+ 1) + 1]3

The left-hand side of (12) can be written as

2+ 32+ 52++ (2k1)2+ [2(k+ 1)1]2:

In view of (11), this simplies to:

12+ 32+ 52++ (2k1)2+ (2k+ 1)2

7.1 + 5 + 9 + 13 ++ (4n3) = 2n2n

Proof:

Assuming the statement is true forn=k:

1 + 5 + 9 + 13 ++ (4k3) = 2k2k;(13)

1 + 5 + 9 + 13 ++ [4(k+ 1)3] = 2(k+ 1)2(k+ 1):(14)

The left-hand side of (14) can be written as

1 + 5 + 9 + 13 ++ (4k3) + [4(k+ 1)3]:

In view of (13), this simplies to:

8.2 + 23+ 25++ 22n1=2(22n1)3

Proof:

Forn= 1, the statement reduces to 2 =2(221)3

Assuming the statement is true forn=k:

3+ 25++ 22k1=2(22k1)3