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Series and Induction

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Introduction

image00.png

INTRODUCTION

The definition of sequence is a list of numbers written in a particular order. These numbers are called the terms of the sequence. For example, 2,4,6,8 are the first four terms of positive even integers. When the sum of the terms in a sequence is taken, we will get something called a series. For example, 2+4+6+8+… is a series. The terms in a sequence is denoted by Tn where n is the number of the term in question.

In my portfolio there are certain conventions used such as:

‘+’ denotes addition

‘–‘ denotes subtraction

‘x’ denotes multiplication

‘/’ denotes division

‘∑’ denotes summation

image02.png’ denotes equal to

OBSERVATION & ANALYSIS

Q1.  Consider the sequence {an}n=1 where

        a1 = 1 x 2

        a2 = 2 x 3

        a3 = 3 x 4

        a4 = 4 x 5

        .

        .

        .

Find an expression for an, the general term in the sequence.

A1.

Let p = 1

a1 = 1 x 2 = p(p+1)

a2 = 2 x 3 = (p+1)(p+2)

a3 = 3 x 4 = (p+2)(p+3)

a4 = 4 x 5 = (p+3)(p+4)

.

.

Thus,

an = (p+n-1)(p+n)

Since p = 1

Therefore,

an = n(n+1)

Q2. Consider the series Sn = a1 + a2 + a3 +…+ an where ak is defined as above.

       (a). Determine several values of Sk, including S1, S2, S3, …, S6 and note observations.

       (b). Formulate a conjecture for a general expression for Sn.

       (c). Prove your conjecture by induction.

       (d). Using the above result, calculate 12 + 22 + 32 + …+ n2.

A2.

(a).

Sn = a1 + a2 + a3 + …………. + an

S1 = a1 = 1 x 2

 = 2

S2 = a1 + a2 = (1 x 2) + (2 x 3)

                = 2 + 6

                = 8

S3 = a1 + a2 + a3 = (1 x 2) + (2 x 3) + (3 x 4)

                       = 2 + 6 + 12

                    = 20

S4 = a1 + a2 + a3 + a4 = (1 x 2) + (2 x 3) + (3 x 4) + (4 x 5)

                            = 2 + 6 + 12 + 20

                            = 40

S5 = a1 + a2 + a3 + a4 + a5 = (1 x 2) + (2 x 3) + (3 x 4) + (4 x 5) + (5 x 6)

                                = 2 + 6 + 12 + 20 + 30

                                = 70

S6 = a1

...read more.

Middle

, …, T6 and note observations.

       (b). Formulate a conjecture for a general expression for Tn.

       (c). Prove your conjecture by induction.

       (d). Using the above result, calculate 13 + 23 + 33 + …+ n3.

A3.

(a).

Tn = 1x2x3 + 2x3x4 + 3x4x5 +….+ n(n + 1)(n + 2)

T1 = 1 x 2 x 3

     = 6

T2 = (1 x 2 x 3) + (2 x 3 x 4)

     = 6 + 24

     = 30

T3 = (1 x 2 x 3) + (2 x 3 x 4) + (3 x 4 x 5)

     = 6 + 24 + 60

     = 90

T4 = (1 x 2 x 3) + (2 x 3 x 4) + (3 x 4 x 5) + (4 x 5 x 6)

     = 6 + 24 + 60 + 120

     = 210

T5 = (1 x 2 x 3) + (2 x 3 x 4) + (3 x 4 x 5) + (4 x 5 x 6) + (5 x 6 x 7)

     = 6 + 24 + 30 + 120 + 210

     = 420

T6 = (1 x 2 x 3) + (2 x 3 x 4) + (3 x 4 x 5) + (4 x 5 x 6) + (5 x 6 x 7) + (6 x 7 x 8)

     = 6 + 24 + 60 + 120 + 210 + 336

     = 756

A graph between Tn and n showing how Tn changes with a change in n.

image01.png

(b).

Tn = (1 x 2 x 3) + (2 x 3 x 4) + (3 x 4 x 5) +…+ n(n + 1)(n + 2)

Tn = ∑ n(n + 1)(n + 2)

     = ∑ (n3 + 3n2 + 2n)

     = ∑ n3 + 3∑ n2 + 2∑n

We know that n4 – (n-1)4 = n4 – n4 + 4n3 – 6n2 + 4n – 1

                                = 4n3 – 6n2 + 4n – 1

Thus,

            14 – 04   = 4(1)3 – 6(1)2 + 4(1) – 1

            24 – 14   = 4(2)3 – 6(2)2 + 4(2) – 1

            34 – 24   = 4(3)3 – 6(3)2 + 4(3) – 1

                        .

                        .

                        .

         + n4 – (n-1)4 = 4(n)3 – 6(n)2 + 4(n) – 1

        n4 – 04 = 4(13 + 23 +…+ n3) – 6(12 + 22 +…+ n2) + 4(1 + 2 +…+ n) – n

image02.png n4 = 4∑n3 – 6∑n2 + 4∑n – n

image02.png n4 = 4∑n3 – [6n(n + 1)(2n + 1)/6] + [4n(n + 1)/2] – n

image02.png n4 = 4∑n3 – [n(n + 1)(2n + 1)] + [2n(n + 1)] – n

image02.png ∑n3 = [n4 + n + (2n3 + 3n2 + n) – (2n2 + 2n)]/4

               = (n4 + n + 2n3 + 3n2 + n – 2n2 – 2n)/4

               = (n4 + 2n3 + n2)/4

               = [n2(n2 + 2n + 1)]/4

        ∑n3 = [n2(n + 1)2]/4

Hence,

Tn = ∑n3 + 3∑n2 + 2∑n

     = [n2(n + 1)2/4] + [3n(n + 1)(2n + 1)/6] + [2n(n + 1)/2]

     = [n2(n + 1)2/4] + [n(n + 1)(2n + 1)/2] + [n(n + 1)]

     = [n2(n + 1)2 + 2n(n + 1)(2n + 1) + 4n(n + 1)]/4                (taking 4 as the LCM)

     = n(n + 1)[n(n + 1) + 2(2n + 1) + 4]/4

     = n(n + 1)(n2 + n + 4n + 2 + 4)/4

     = n(n + 1)(n2 + 5n + 6)/4

     = n(n + 1)(n + 2)(n + 3)/4

Tn = n(n + 1)(n + 2)(n + 3)/4

(c).

1x2x3 + 2x3x4 + 3x4x5 +…+ n(n + 1)

...read more.

Conclusion

k-2 +…. + n(-1)k+1

                                      k+1

Hence,

1k + 2k + 3k +…+ nk = nk+1+ [k(k+1)/2] ∑nk-1 – [k(k+1)(k-1)/6] ∑nk-2 +…. + n(-1)k+1

                                                              k+1

CONCLUSION

After going through my portfolio, I hope I have satisfied the basic needs required and have presented the portfolio well. I have learnt a lot of things while doing this investigation. Firstly, my knowledge about sequences and series have been broadened and I am now craving for more knowledge on general formulas of these sequences. There is a lot to learn from this portfolio. We can use the general formulas of Sn, Un and Tn in our everyday lives. We should observe the beauty of the portfolio, the beauty of the patterns, the way they progress at every stage. I was so inspired that I looked up on the internet for more methods to solve this portfolio. Infact I even studied the binomial theorem and the pascal’s triangle deeply as it interested me. While surfing on the internet I found something very interesting. The method I used to find the conjectures of Sn, Un and Tn has a name and is called the Telescopic Method.

REFERENCE:

http://www.xula.edu/math/Research/Colloquium/Past/Colloquium20030320-Paper-SahooP.pdf

http://www.mathpages.com/home/kmath279.htm

http://mathworld.wolfram.com/PowerSum.html

http://mathworld.wolfram.com/BinomialTheorem.html

All graphs have been made with the help of Microsoft Excel 2003

All major calculations have been done with the help of Graphic Display Calculator TI-84 Plus.


[1]k+1C1 = (k+1)!/[(k+1-1)!x1!] = (k+1)k!/k! = k+1

k+1C2 = (k+1)!/[(k+1-2)!x2!] = k(k+1)(k-1)!/[(k-1)!x2] = k(k+1)/2

k+1C3 = (k+1)!/[(k+1-3)!x3!] = k(k+1)(k-1)(k-2)!/[(k-2)!x6] = k(k+1)(k-1)/6

...read more.

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