Example: Consider the sequence {a i = i} i=1. Starting with i = 1, since a i = i,

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1 Sequeces: Defiitio: A sequece is a fuctio whose domai is the set of atural umbers or a subset of the atural umbers. We usually use the symbol a to represet a sequece, where is a atural umber ad a is the value of the fuctio o. Ituitively, a sequece is just a ordered list of (possibly ifiitely may umbers. Each umber i a sequece is a term of the sequece. We usually use the letter i as the idex, ad a i is the i-th term of the sequece. A sequece may be fiite or ifiite. If a sequece is fiite, we sometimes write {a 1, a, a, a 4,... a } to represet the sequece; If a sequece is ifiite, we write {a 1, a, a,... } or {a i }. The otatio {a i } implies that we have a sequece whose first term is a 1, the secod term is a, the third term is a...etc. The idex i starts from 1 (or ay other positive iteger ad icreases by 1 each time to represet each subsequet term i the sequece. A sequece ca be represeted by a formula expressed as a expressio i i or i. If a sequece has a patter we ca also write the first few terms of the sequece ad assume that the patter cotiues ad let the reader figure out the values of the subsequet terms. A sequece ca also be defied recursively, where we assig a value to the first (or first few terms of the sequece, ad the value of each term i the sequece is the defied by oe or more of the precedig terms. We ca also describe a sequece verbally if there s o obvious formula or patter that we ca use to express the sequece. Example: Cosider the sequece {a i 6}. Startig with i 1, sice a i 6, so a 1 6 is the first term of the sequece. If i, the a 6. If i, the a 6. The value of a i is always the same value, so we have the sequece of costat terms: {a i 6} { 6, 6, 6, 6,... } Example: Cosider the sequece {a i i}. Startig with i 1, sice a i i, so a 1 1 is the first term of the sequece. If i, the a. If i, the a. Cotiue i this fashio, we obtai the sequece of positive itegers: {a i } {1,,, 4,... } Example: Cosider the sequece { a i i }. Startig with i 1, sice a i i, so a 1 1 is the first term of the sequece. If i, the a 1 is the secod term. If i, the a 6 is the third term. If i 4, the a is the fourth term. Cotiue i this fashio, we obtai the followig sequece: {a i } {, 1, 6, 1,,, 46,... }

2 Example: Cosider the sequece a i i i + 1, so a { a i i }. Startig with i 1, sice i + 1 is the first term of the sequece. If i, the a + 1 is the secod term. If i, the a is the third term. If i 4, the a is the fourth term. Cotiue i this fashio, 5 { 1 we obtai the followig sequece: {a i },, 4, 4 5, 5 } 6,... Example: Cosider the sequece: {, 4, 6, 8,... }. Assumig the patter cotiues, this is the sequece of positive eve itegers. We ca also represet this sequece usig a formula: {a i i} or {i} Example: {1, 1, 1, 14, 15 }, Assumig the patter cotiues, this is a sequece whose terms alterate i sig. It ca be expressed by a i ( 1i+1 i Example: Cosider the sequece defied by a i a i 1 +a i ; a 1 1, a 1. This is a sequece defied recursively. The value of the first two terms is assiged to be 1 ad 1, the the value of ay ext terms is defied to be the sum of the precedig two terms. From the defiitio of this sequece, we have: a a +a 1 1+1, a 4 a + a + 1, a 5 a 4 + a + 5, a 6 a 5 + a , a 7 a 6 + a Cotiue i this patter, we obtaied the followig sequece of umbers: {1, 1,,, 5, 8, 1, 1, 4, 55, 89, 144, } This sequece is called the sequece of Fiboacci umbers. Notice that whe defiig a sequece recursively, oe ca chage the iitial values of the sequece while keepig the same relatioship betwee the terms, ad we will ed up with a sequece with differet terms. I the previous example, if we defie: a i a i 1 + a i ; a 1, a 4, we ed up with this sequece: {, 4, 7, 11, 18, 9, 47, 78, 15, } Example: Defie a sequece recursively by: a (a 1 ; a 0 1 We obtai the terms of this sequece as: a 1 1(a 0 1(1 1, a (a 1 (1, a (a ( 6, a 4 4(a 4(6 4, a 5 5(a 4 5(4 10. This sequece is used ofte i mathematics ad is amed the factorial, ad we use the symbol!, to represet, we write:! ( 1! 0! 1 Notice that by defiitio, 0! 1, 1! 1(0! 1(1 1,! (1! (1,

3 ! (! ((1! ((1, 4! 4(! 4(((1. I geeral,! ( 1( (((1 So we ca defie factorial i a differet way: For ay o-egative iteger,! 1 if 0! ( 1( (((1 if 1 Usig this defiitio, we ca evaluate: 7! 7(6(5(4((( ! 11(10(9(8(7(6(5(4((( Example: Cosider the sequece defied by: a i the i-th prime umber. This is the sequece {,, 5, 7, 11,... }. This sequece caot be expressed as a expressio i i, but is well-defied. Example: The sequece {1, 1, 1,... } is a sequece defied by a i 1. Example: The sequece {i } is the sequece of positive perfect cubes, {a i } {1, 8, 7, 64, 15,... } { 1 1 Example: The sequece is the sequece: i + 1, 1 5, 1 10, 1 } { 1 Example: Fid a expressio i i for the sequece 4, 9, 16, 4 } 5,... As: For each fractio, the deomiator is the square of a umber oe greater i tha the umerator, so we may use: a i (i + 1 { Notice that we may also use: a i i 1 }. For ay sequece, i does ot have i i to start at 1. Example: Fid a expressio i i for the sequece {,, 4, 54, 65 },... As: This is a sequece where the deomiator is oe less tha the umerator. To make the terms alterate i sig, we use a power of 1: ( i + 1 a i ( 1 i+1, i 1 i Sigma Notatio Suppose we have a sequece {a i } {a 1, a, a,... }, ofte times we wat to add some or all of the terms of the sequece to fid the sum. Istead of writig a 1 + a + a + + a i + every time, we use the sigma otatio, Σ, to represet the sum of all these terms:

4 Defiitio: a i a 1 + a + a + + a 1 + a Example: Cosider: 8 i. This meas that a i i, we start with i 1, so a 1 1, the icreases i by 1 each time util we get to 8, we have: 8 i Example: Cosider: 11 i. This meas that a i i, we start with i 1, so a 1 1 1, the icreases i by 1, we get a 4, the icreases i by 1 agai, so a 9. Cotiue i this patter util i 11, we have: 11 i Example: Cosider: 1 i5 (i + 1. This meas that a i (i + 1, but this time to get the sum, we start with i 5, so a 5 ( , the icreasmet i by 1, we have a 6 ( Cotiue i this fashio util i 1, we have: 1 i5 (i Example: Cosider: 9 4. This meas that a i 4 for all i. I other words, a 1 4, a 4,...etc., so Some formula ivolvig sigma that would be useful to kow: If c is a costat, c c (we are addig the same costat, c, for may times, the result is times c ca i c a i (this is just the distributive property (a i + b i a i + b i (commutative property of additio

5 (a i b i i a i (( + 1 i (( + 1( [ ] (( + 1 i b i

6 Arithmetic Sequece A arithmetic sequece is a sequece where the differece betwee ay two cosecutive terms is a costat. This costat is called the commo differece. We ususally use the letter d to represet the commo differece. Example: {, 7, 11, 15, 19,, }. This is a arithmetic sequece, the commo differece betwee ay two cosecutive terms is the umber d 4. Example: {8, 5,, 1, 4, 7, 10, 1, } This is a arithmetic sequece, the commod differece is the umber d. Example: The sequece a + 1 is a arithmetic sequece, if we look at the differece betwee ay two cosecutive terms a ad a +1, we get: a +1 a ((+1+1 (+1 (++1 (+1 (+ (+1. The commod differece is d. The sequece defied by the formula is: {, 5, 7, 9, 11, } Example: Fid the -th terms of the arithmetic sequece where the first term is a 1 ad commod differece is d. By defiitio of a arithmetic sequece, the ext term of the sequece ca be obtaied by addig the commo differece d to the precedig term, so we ca obtai the term a by: a a 1 + d We ca similarly obtai the third term, a, by: a a + d (a 1 + d + d Cotiuig the patter gives us: a 4 a + d (a + d + d (a 1 + d + d a 1 + d a 5 a 4 + d (a 1 + d + d a 1 + 4d. a a 1 + ( 1d We obtaied the followig formula: For a arithmetic sequece, if the first term is a 1 ad the commod differece is d, the the th term a is give by: a a 1 + ( 1d Example: Give a arithmetic sequece, {a 1, a, a, a 4,, a i }, fid the sum of

7 the first terms of the sequece, i other words, fid: Sice {a i } is a arithmetic sequece, the i th term of the sequece ca be expressed as a i a 1 + (i 1d, where d is the commo differece, so we write the above as: a i a 1 + (i 1d a 1 + (i 1d ( ((a 1 + d (i 1 ((a 1 + d 1 i ( ( 1( ((a 1 + d ((a 1 + d (( 1( 1 ( [a 1 + d( 1] We just discovered the followig formula: The sum of the first terms of a arithmetic sequece, with first term a 1 ad commo differece d, is give by: S [a 1 + d( 1] Sice i a arithmetic sequece, the term, a, ca be expressed as a a 1 + d( 1, we ca chage the formula to be i terms of the first ad th term: S [a 1 + d( 1] [a 1 + a 1 + d( 1] [a 1 + (a 1 + d( 1] [a 1 + a ] Formula: The sum of the first terms of a arithmetic sequece, where the first term is a 1 ad the term is a, is give by: a i

8 S (a 1 + a Example: Fid a formula for the give sequece, the fid the value of the 5th term of the sequece, ad fid the sum of the first 40 terms. { 7,,, 8, 1, 18,, } The differece betwee each term ad the ext is the costat 5, so this is a arithmetic sequece, with a 1 7 ad d 5, usig the formula, we have: a 5 a 1 + d( (5(4 11 Sice we kow the commo differece ad the first term, we ca use the formula: S [a 1 + d( 1] to fid the sum of the first terms: S [( 7 + 5(40 1] 0 [ (9] 60 Example: A sequece is give by {a 4 6}. Fid the sum of the first 1 terms of this sequece. If we take the differece betwee ay two terms, a 1, a of this sequece, we get: a a 1 (4 6 (4( ( The differece betwee ay two terms of this sequece is a costat, 4. This is a arithmetic sequece with commo differece d 4. The value of the first term is a 1 4(1 6. We ca ow apply the formula: S [(( + (4(1 1] [ ] 1798 Example: Fid the sum of the first 17 terms of the followig sequece: {, 7 } , 5,, 8,, 11,, 14, The differece betwee ay cosecutive two terms is, so this is a arithmetic sequece with commo differece d We use the formula: S [ ( + ] ( [4 + 4] 8 Example: Fid the sum of the first 5 terms of a arithematic sequece {a } if a 10 1 ad a As: We eed to kow the the value of the first term a 1 ad the commo differece d before we ca aswer the questio, we use the formula: a a 1 + ( 1d, to set up equatios to solve for a 1 ad d: a 10 a 1 + (10 1d 1 a 1 + 9d 1

9 a 16 a 1 + (16 1d 49 a d 49 We have a system of two equatios with two ukow. Solvig this system gives us: a 1 + 9d 1 a d 49 6d 6 d 6 a 1 + 9( 6 1 a 1 41 Now we ca use the formula to fid the sum of the first 5 terms: S [ (41 + ( 6(5 1] [8 + ( 04] 15

10 Geometric Sequece Defiitio: A sequece {a } is a geometric sequece if the ratio betwee a term ad its precedig term is a (o-zero costat. This costat is called the commo ratio of the sequece ad is usually represeted by r. I other words, r a a 1 Example: Cosider the sequece {, 4, 8, 16,, 64, }. The ratio betwee ay term of the sequece ad its precedig term is the umber, so this is a geometric sequece. The commo ratio is the umber r. { Example: The sequece 4, 4, 4 9, 4 7, 4 81, 4 } 4,. The ratio betwee ay term ad its precedig term is the umbe, so this is a geometric sequece. The commo ratio is r 1 Example: Fid the expressio for the th term of a geometric sequece if the first term is a 1 ad the commo ratio is r. As: By defiitio, r a a ra 1. I other words, to obtai the ext a 1 term i a geometric sequece, we multiply the commo ratio to the previous term, we have: a a 1 r a a r (a 1 r(r a 1 r a 4 a r (a 1 r r a 1 r. Lookig at the patter, we ca deduce a formula for the th term of a geometric sequece: Formula: If a geometric sequece has first term a 1 ad commo ratio r, the its term, a, is give by: a a 1 r 1 Example: Fid the 4th term of the sequece: { 4 9, 1, 1 4, 16, 9 64, 7 56, 81 } 104, As: The ratio betwee each term of this sequece ad the previous term is the same, 4, so this is a geometric sequece with r 4, a 1 4, the value of the 9 4th term is give by: a a 1 r 1 a 4 ( 4 9 ( 4 1 ( (

11 Example: Fid the sum of the first terms of a geometric sequece {a } where the first term is a 1 ad the commo ratio is r. As: To add the first terms of a geometric sequece with first term a 1 ad commo ratio r, we eed to fid the followig sum: S a 1 + a 1 r + a 1 r + a 1 r + a 1 r a 1 r 1 If we factor the commo factor a 1 i the expressio we have: S a 1 +a 1 r+a 1 r +a 1 r +a 1 r 4 + +a 1 r 1 a 1 ( 1 + r + r + r + r r 1 Cosider the fuctio f(x x 1. Notice that f( Sice f(1 0, the factor theorem tells us that x 1 is a factor of f(x. If we divide f(x by x 1, we get the followig: f(x x 1 (x 1(x 1 + x + x + x + x + x + 1 I other words, x 1 + x + x + x + x + x + 1 x 1 x 1 Replace x with r we ca ow complete the above formula: The sum of the first terms of a geometric sequece, if the first term is a 1 ad the commo ratio is r, is give by: ( S a r + r + r + r r 1 ( r 1 a 1 Example: Fid the sum of the first 8 terms of the sequece: { ( } a 5 As: If we divide ay term by its precedig term, we get: a ( 5 a 1 ( This is a geometric sequece with commo ratio r 5 ad first term a 1 ( 1 6. Usig the formula, the sum of the first 8 terms is: 5 5 ( r 8 (( 1 S 8 a ( ( ( Example: Fid a expressio for the term of the geometric sequece {a } if a 4 ad a As: We eed to first fid the first term, a 1, ad the commo ratio, r. We use

12 the formula to set up two equatios: a a 1 r 1 a 6 a 1 r a a 1 r 1 a a 1 r 1 4 We get two equatios: a 1 r a 1 r 4 Divide the two equatios give: r 7 r Solvig for a 1 i the equatio gives: a 1 ( 4 9a 1 4 a So the expressio for the sequece is: a 4 9 ( 1

13 Priciple of Mathematical Iductio: Suppose a statemet is made about some (or all atural umbers. Suppose we ca prove the followig: 1. The statemet is true for 1.. If the statemet is true about some atural umber, it is also true for the ext atural umber, + 1. The the statemet is true for all atural umbers. We use the priciple of mathematical iductio to prove mathematical theorems ad formulas about atural umbers. Notice that the first coditio tells us that the statemet is true for the very first umber, 1. The secod coditio assures us that, if a statemet is true for a atural umber, it must be true for the ext umber. So sice the statemet is true for 1, it must be true for the ext umber, ; ad usig the secod coditio agai, sice the statemet is true for, it must be true for, ad cotiuig i this argumet, we ca see that the statemet must be true for all atural umbers. I usig mathematical iductio to prove a statemet about all atural umbers, we first prove that the statemet is true for the first umber, 1. The we ca assume that the statemet is true for (this is called the iductio hypothesis, if we ca the prove, from this assumptio, that the statemet is also true for + 1, the we have proved the statemet is true for all atural umbers. Example: Prove the formula: i ( 1 + (( + 1 As: We use mathematical iductio o. We first eed to prove the the formula is true for 1, but this is trivial: The left had side is: 1 i 1 (1(1 + 1 The right had side is 1 Now we assume that the formula is true for. I other words, we assume ow that (( ( 1 + is true.

14 We must ow prove that the formula is true for + 1. I other words, we must prove this formula is true: ( + 1(( ( 1 + ( + ( + 1 We will use the assumptio that the formula is true for. Accordig to the formula: (( + 1 [ ( 1 + (] + ( ( + 1 (( + 1 ( + 1 (( ( ( + 1 [ + ] ( + 1 [( ] We showed the formula is also true for + 1 if we assume that it is true for, so we have proved the formula is true for all atural umbers by the priciple of mathematical iductio. Example: Use mathematical iductio to prove ( the sum of a geometric sequece r formula: a + ar + ar + ar + + ar 1 1 a As: We first show that the formula is true for 1. ( r 1 1 If 1, left had side is a. Right had side is a a We ow assume the formula is true for. We eed to prove that it is true for + 1. I other words, we eed to prove that: ( r a + ar + ar + ar + + ar 1 + ar ( a By assumptio, the formula is true for, so o the left had side we have: [ a + ar + ar + ar + + ar 1] + ar (+1 1 [ a ( r ] [ 1 + ar (+1 1 a a(r 1 + ar ( a(r 1 + ar ( a [ r 1 + r +1 r ] ( r ] 1 + ar a(r 1 + ar a(r 1 + ar ( a [(r 1 + r (] a [ r +1 1 ] ( r +1 1 a

15 Suppose, k are both o-egative itegers ad k, we defie ( (read choose k to be: k (! k k!( k! ( 8 8! Example:!(8! 8!! 5! ( 11 11! Example: 4 4!(11 4! 11! ! 7! 4 ( 1 1! Example: 1 1!(1 1! 1! 1! 0! 1! 1! 1 1 ( 7 7! Example: 6 6!(7 6! 7! 6! 1! 7 ( 14 14! Example: 0 0!(14 0! 14! 1 14! 1 I geeral, ( 1 ( 1 ( 1 0 ( 1 ( ( k k Formula: ( ( + k k 1 Proof: ( + k ( k 1 ( + 1 k! k!( k! +! (k 1!( k + 1!! k!( k! +! (k 1!( (k 1!! ( k + 1 k!( k! ( k + 1 +! (k 1!( k + 1! k k

16 ! ( k + 1 k!( k + 1 ( k! +! k k (k 1!( k + 1!! ( k + 1 k! ( k + 1! +! k! ( k + 1 +! k k! ( k + 1! k! ( k + 1! (! [ + 1] k! ( k + 1! ( + 1! + 1 k! ( + 1 k! k ( ( ( For example, + For example, 6 + Biomial Theorem: 5 6 ( + 1 Let be a positive iteger, the (a + b ( a a 1 b + ( a b + a b + 1 ab 1 + a b + + ( b Proof: We will use mathematical iductio o. If 1, this is trivial.! [( k k] k! ( k + 1! Now assume that the formula is true for, we eed to show that it is also true for + 1. (a + b +1 (a + b (a + b By the iductio assumptio, the formula is true for, so (a + b ( ( ( ( a + a 1 b + a b + a b ( a b + 1 Therefore, (a + b (a + b [( ( ( a + a 1 b + a b ( a b + ( a a b + a 1 b + ab 1 + ab a b + ( b a b + + ( ] b (a + b a b a b 1 + ( ab

17 ( ( ( + a b + a 1 b + a b ( ( ( a b 1 + ab + b +1 1 ( ( ( a +1 + a (+1 1 b + a (+1 b + ( 0 ( ab + b ( a b a (+1 b + Startig with 0, the coefficiets of (a + b is the etries i the th row of the Pascal s Triagle. Example: Expad (x + y 6 As: Usig the biomial theorem, ( ( ( (x+y 6 x 6 + x 5 y + x 4 y ( 6 x y + x 6 + 6x 5 y + 15x 4 y + 0x y + 15x y 4 + 6xy 5 + y 6 ( 6 x y ( 6 xy ( 6 y 6 6