• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Series and Induction

Extracts from this document...

Introduction 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 subtraction

‘x’ denotes multiplication

‘/’ denotes division

‘∑’ denotes summation ’ 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

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. (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 n4 = 4∑n3 – 6∑n2 + 4∑n – n n4 = 4∑n3 – [6n(n + 1)(2n + 1)/6] + [4n(n + 1)/2] – n n4 = 4∑n3 – [n(n + 1)(2n + 1)] + [2n(n + 1)] – n ∑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)

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.

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

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Extended Essay- Math

!`(dï¿½ï¿½ ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½Gï¿½ÆÇoï¿½I`ï¿½-(tm)(tm)ï¿½ï¿½ï¿½ï¿½PHï¿½ï¿½ï¿½ï¿½U|G_Éï¿½ï¿½ï¿½Kbbï¿½sï¿½=ï¿½ï¿½ï¿½/ uï¿½Ü¹ï¿½)))999ï¿½ï¿½ï¿½ï¿½ï¿½ Aï¿½rï¿½ï¿½sÕªU(c)(c)(c)>\ï¿½rï¿½ï¿½ï¿½"BHï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½Jï¿½iï¿½ï¿½ï¿½~ï¿½&" .ï¿½:uï¿½\ï¿½ï¿½V!ï¿½\ï¿½ï¿½sï¿½ï¿½-qpp(Wï¿½ï¿½\$ï¿½@ï¿½|(ï¿½Qï¿½CÙ¸qï¿½--[-<xpï¿½ï¿½á°°ï¿½ï¿½'- ï¿½ï¿½(c)Pï¿½ï¿½ï¿½ï¿½e+ï¿½ï¿½5kï¿½L(tm)rï¿½ÆZï¿½jï¿½Mï¿½gB@ È"ï¿½bï¿½<vï¿½Ø¨Qï¿½>ï¿½ï¿½"mÛ¶ï¿½""WXï¿½ï¿½'#ï¿½ï¿½ï¿½}*ï¿½,\$ 4(??ï¿½W'ï¿½3/4"X!@S-VHZOnn(r)ï¿½ï¿½-ï¿½&gΜï¿½ï¿½ï¿½3/4wï¿½vï¿½kï¿½(r)ï¿½Nï¿½0ï¿½ï¿½ï¿½vvvï¿½ Vl"fÍUï¿½ï¿½ ï¿½ï¿½ï¿½ï¿½S|m1/2{ï¿½vttï¿½\ï¿½o1/2ï¿½ï¿½ï¿½Åï¿½m}ï¿½ï¿½Wï¿½ï¿½ï¿½ï¿½ï¿½VVVï¿½?^1/2zu...Nq5S)B ÈR8"ï¿½ï¿½CA{ï¿½ï¿½ï¿½ï¿½ï¿½"ï¿½ï¿½ (r)ï¿½ =yï¿½iÓ¦ï¿½+ï¿½ï¿½"ï¿½3/4ï¿½ï¿½Ø³ï¿½B(ï¿½(r)ï¿½r Fï¿½ï¿½ï¿½ V ï¿½Fï¿½ o#y #@"ï¿½`ï¿½h"oï¿½ï¿½1'-P0D( V ï¿½Fï¿½ o#y #@"ï¿½`ï¿½-Zqqï¿½ï¿½ï¿½? ï¿½`5ï¿½ï¿½ï¿½>>>^ï¿½"|&ï¿½ï¿½Caï¿½|ï¿½ï¿½Pï¿½&'ï¿½'ï¿½^1/2zï¿½ï¿½ï¿½~~~ï¿½z "ï¿½ {]ï¿½%ï¿½&8dï¿½'R8ï¿½sï¿½ï¿½9M =ï¿½Aï¿½D ï¿½ï¿½ï¿½Y9jPï¿½\$666yyyJ-ï¿½dS &J(ï¿½+ï¿½ï¿½Ñ¨ï¿½-ï¿½ ï¿½ ^"vMï¿½Ð°aï¿½Nï¿½8Aï¿½zPï¿½"ï¿½[Gï¿½-E@yD?ï¿½Ø±#,ï¿½]^ï¿½'''ï¿½ï¿½ï¿½uï¿½Ô±ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½zï¿½Aï¿½ (tm) Bï¿½ï¿½ï¿½ï¿½-# Ò {qï¿½(r)]z-×³`eï¿½uï¿½ï¿½( ï¿½a ï¿½#"'""]>hÊ£7ï¿½Ó¦Mï¿½-tÑ¢E3/43/43/4Xpï¿½ ï¿½ï¿½vÊ£ï¿½|Tï¿½0-D(zcï¿½ï¿½}ï¿½@ï¿½Yï¿½fDDD@@ï¿½vy"mL(ï¿½D Bï¿½[Ñ³gÏ¶ï¿½ï¿½D<ï¿½Í7ï¿½ï¿½"ï¿½yï¿½ï¿½1ï¿½ï¿½Íï¿½(c)K;"ï¿½ï¿½+z Tï¿½ï¿½~ï¿½-**jÙ²ï¿½K-P 7ï¿½a+ï¿½ï¿½ ï¿½ ï¿½ï¿½ @NYï¿½t ï¿½ï¿½"ï¿½ï¿½@ ï¿½ï¿½Vï¿½ï¿½""rjAï¿½ï¿½Tï¿½ï¿½ÚµkÏ9ï¿½q 9ï¿½(r)[ï¿½Nï¿½ï¿½"ï¿½P54ï¿½ï¿½^ï¿½t`ï¿½TJï¿½ï¿½<b0ï¿½2ï¿½@ï¿½...1/2ï¿½Pï¿½cJ5rï¿½ï¿½P8Qï¿½Iï¿½ Z" N Bï¿½DQ\$&!ï¿½D(<hï¿½d\$8Aï¿½...E'ï¿½"ï¿½ï¿½ %''ï¿½"NEb<

2. ## How many pieces? In this study, the maximum number of parts obtained by n ...

2. 3. Conjecture by technology: Qn = (1/24)n 4 - (1/12)n3 + (11/24)n2 + (7/12)n + 1 Prove Conjecture: n ? (1/24)n 4 - (1/12)n3 + (11/24)n2 + (7/12)n + 1 = (2 + 4 + 8 + 16 + 31 + .....+n) n=1 Check n = 1: (1/24)(1)

1. ## Infinite summation portfolio. A series is a sum of terms of a sequence. A ...

It's due to the fact that the decimal number we've taken is less than 1, and it could not be a limitation for the statement that seems to be again Sn = ax. To be sure we can check the statement with choosing a square root for x.

2. ## Infinite Summation Portfolio. I will consider the general sequence with constant values

is this: Graph 1: Shows the relation between and in We can observe that the values increase exponentially until they reach the number 2. At this point they stabilize. Hence, we can conclude that this sequence is convergent. In other words, it is approaching a definite limit as more of its terms are added.

1. ## Infinite Summation - In this portfolio, I will determine the general sequence tn with ...

Consider the sequence where and : Using GDC, I will calculate the sums S0, S1, S2, ..., S10: S0 = t0 = 1 S1 = S0 + t1 = 1 + = 2.945910 S2 = S1 + t2 = 2.945910 + = 4.839193 S3 = S2 + t3 = 4.839193

2. ## Infinite Summation

Ä Uï¿½ ï¿½?ï¿½!ï¿½ï¿½"!ï¿½!ï¿½!ï¿½Xï¿½puB Lï¿½ 2ï¿½'ï¿½;ï¿½x ï¿½,PyWï¿½Bï¿½Rï¿½Dï¿½ï¿½L ï¿½y ~P^Qï¿½ï¿½@ï¿½ï¿½A1dï¿½-ï¿½Aï¿½(y!Lï¿½ï¿½ Mï¿½S "O>ï¿½j9@ï¿½L ï¿½< \ï¿½Bï¿½yLmvLï¿½LXTï¿½ Qï¿½ï¿½PK !ï¿½ï¿½fÈ¨ï¿½word/media/image52.pngï¿½PNG IHDR ï¿½M\ï¿½%iCCPICC Profilex..."MHaï¿½ï¿½ï¿½ï¿½ï¿½-ï¿½ï¿½\$T& Rï¿½+Sï¿½eï¿½L bï¿½}wï¿½gï¿½(tm)ï¿½-E""ï¿½uï¿½.VDï¿½ï¿½Nï¿½Cï¿½:D(tm)uï¿½ ï¿½E^"ï¿½ï¿½;""cT3/403ï¿½yï¿½ï¿½ï¿½|1/2ï¿½Uï¿½Rï¿½cE4`ï¿½ï¿½"ï¿½ÞvztLï¿½Uï¿½F\)ï¿½s:ï¿½ï¿½(c)ï¿½ï¿½kï¿½-iYj"ï¿½ï¿½6|"v(tm)P4*wd>,y<ï¿½ï¿½'/ï¿½<5g\$(c)4ï¿½!7ï¿½Cï¿½Nï¿½-ï¿½ï¿½lï¿½ï¿½Cï¿½ï¿½Tï¿½S"3-q";ï¿½-E#+c> ï¿½vÚ´ï¿½ï¿½=ï¿½SÔ°ï¿½ï¿½79 Ú¸ï¿½@ï¿½-`Óï¿½mï¿½-ï¿½vï¿½Ulï¿½5ï¿½ï¿½`ï¿½Pï¿½=ï¿½ï¿½Gï¿½ï¿½ï¿½jï¿½ï¿½)ï¿½kï¿½P*}ï¿½6ï¿½~^/ï¿½~ï¿½.ï¿½~ï¿½a ï¿½ï¿½ï¿½2 nï¿½×²0ï¿½%ï¿½ï¿½fï¿½ï¿½ï¿½ï¿½ï¿½ï¿½|U ï¿½ï¿½9ï¿½lï¿½ï¿½7?ï¿½ï¿½ï¿½j`ï¿½'ï¿½l7ï¿½ï¿½ï¿½"ï¿½tï¿½iï¿½ï¿½Nï¿½f]?ï¿½uï¿½h...ï¿½gM ZÊ²4ï¿½ï¿½i(r)ï¿½"[ï¿½&LYï¿½ï¿½_ï¿½xï¿½ {xï¿½Oï¿½ï¿½\$1/4ï¿½ß¬Ì¥S]ï¿½%ï¿½ï¿½Ö§ï¿½ï¿½ï¿½&7ï¿½ï¿½gÌ>r=ï¿½ï¿½*g8`ï¿½(tm)ï¿½ 8rÊ¶ï¿½<(c)ï¿½ï¿½ï¿½ï¿½ï¿½"dï¿½WT'"ï¿½<ï¿½ eLï¿½~.u"A(r)ï¿½=9(tm)ï¿½-ï¿½]ï¿½ï¿½>31ï¿½3'ï¿½X3ï¿½ï¿½ï¿½ï¿½-\$eï¿½}ï¿½ï¿½u,ï¿½ï¿½gm'g...6ï¿½64\$Ñï¿½E zL*LZï¿½_ï¿½jï¿½ï¿½ï¿½_ï¿½ï¿½]1/2Xï¿½ï¿½yï¿½[ï¿½?...Xs ï¿½ï¿½ï¿½Nï¿½ï¿½/ï¿½ ï¿½]ï¿½ï¿½|mï¿½3/4ï¿½(tm)sÏï¿½"k_Wf-ï¿½È¸Aï¿½23/4ï¿½)ï¿½oï¿½ï¿½ z-diï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½2ï¿½|mÙ£ï¿½ï¿½j|5Ô¥ejï¿½8ï¿½ï¿½(r)eï¿½Eï¿½ï¿½7ï¿½ï¿½[ï¿½ï¿½ï¿½Qï¿½|ï¿½IM%×²ï¿½xf)ï¿½|6\ kï¿½ï¿½"`Ò²"ï¿½ä.<kï¿½ï¿½Uï¿½}jï¿½ï¿½ M=ï¿½ï¿½"mjßï¿½ï¿½Üï¿½ï¿½ ï¿½ï¿½eï¿½)ï¿½`cï¿½ï¿½ï¿½IWfï¿½ï¿½ï¿½ï¿½/ï¿½ï¿½ï¿½^a ï¿½44ï¿½Mï¿½ï¿½ï¿½i ï¿½ï¿½6pï¿½"ï¿½ï¿½_ï¿½ Þ¡ï¿½ï¿½>IDAT8ï¿½';KQ...ï¿½ï¿½ï¿½)Dï¿½ï¿½ï¿½`'ï¿½`)Vbgkï¿½ï¿½>ï¿½ï¿½ ï¿½tVZ[k' ï¿½ï¿½UDWKA"Xï¿½Yï¿½,ï¿½ï¿½aï¿½3ï¿½3/4ï¿½ï¿½ ï¿½o5ï¿½v~ï¿½)oï¿½"-Xï¿½ï¿½ï¿½Sï¿½ï¿½ï¿½p w&ï¿½ï¿½>Yï­°f_ï¿½}X...3ï¿½Aï¿½ï¿½@ï¿½?Eu^ï¿½ <ï¿½i"d -ï¿½a@ï¿½ mf R ï¿½ï¿½Iï¿½ï¿½.}R6`ï¿½!...9ï¿½ï¿½ï¿½ Éºï¿½,ï¿½&Û£Xï¿½lï¿½uï¿½ï¿½ï¿½x%?ï¿½Tï¿½ï¿½ï¿½ï¿½`ï¿½Ð6ï¿½ï¿½Wï¿½Teï¿½ï¿½ ï¿½Jï¿½ï¿½Sï¿½(r)ï¿½lï¿½Z2"ï¿½ï¿½ï¿½ ï¿½ï¿½ï¿½ï¿½86ï¿½ï¿½×£ï¿½ï¿½Lï¿½ï¿½]ï¿½ ï¿½ï¿½Rï¿½Aï¿½Bï¿½&q^ï¿½ï¿½j@5 ï¿½ï¿½ ...ï¿½-ï¿½"ï¿½(tm)ï¿½ï¿½Aï¿½5Þ¤ 5ï¿½ ï¿½Eï¿½ï¿½ï¿½P(r)

1. ## Math Portfolio: trigonometry investigation (circle trig)

240 -0.866025404 -0.5 1.732050808 260 -0.984807753 -0.173648178 5.67128182 270 -1 -1.83772E-16 5.44152E+15 280 -0.984807753 0.173648178 -5.67128182 300 -0.866025404 0.5 -1.732050808 320 -0.64278761 0.766044443 -0.839099631 340 -0.342020143 0.939692621 -0.363970234 360 -2.4503E-16 1 -2.4503E-16 The number with E is error made by MS Excel calculations.

2. ## Infinite Summation- The Aim of this task is to investigate the sum of infinite ...

will increase too. As x increases the curve slopes up regularly. Now the value of a will be changed to 3 giving T9(3,x) and leaving x to be changed. For this example the values of x will be 1,2,3,4,5,6,7,8,9,10 as in the first example. • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 