Key Guidelines for Low-Income Housing.ptx

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Discrete Mathematics

Sets A set is an unordered collection of different elements. A set can be written explicitly by listing its elements using set bracket. If the order of the elements is changed or any element of a set is repeated, it does not make any changes in the set.

Some Example of Sets A set of all positive integers A set of all the planets in the solar system A set of all the states in India A set of all the lowercase letters of the alphabet Representation of a Set Sets can be represented in two ways − 1)Roster or Tabular Form 2)Set Builder Notation Colours of rainbow. C={V,B,I,G,Y,O,R} C=name of set,V =Elements of set,{}-curly bracket

Cartesian product 1)Let A={1,2,3,4} and B={5,6,7,8}.Find AXB. 2)Let A={ d,a,n,g } and B={ g,e,r }.Find AXB. 3)Let A={ k,i,t,e } and B={ f,r,u,i,t }.Find AXB. Let A={1,2,3} and B={ c,d }.Find AXB. AXB={(1,c),(1,d),(2,c),(2,d),(3,c),(3,d)}

Roster or Tabular Form The set is represented by listing all the elements comprising it. The elements are enclosed within braces and separated by commas . Example 1  − Set of vowels in English alphabet, A={ a,e,i,o,u } Example 2  − Set of odd numbers less than 10, B={1,3,5,7,9 } Set Builder Notation The set is defined by specifying a property that elements of the set have in common. The set is described as A={x:p(x)}A={x:p(x)} Example 1  − The set { a,e,i,o,u }{ a,e,i,o,u } is written as − A={x:x is a vowel in English alphabet}A={x:x is a vowel in English alphabet} Example 2  − The set {1,3,5,7,9}{1,3,5,7,9} is written as − B={x:1≤x<10 and (x%2)≠0} If an element x is a member of any set S, it is denoted by  x∈Sx∈S  and if an element y is not a member of set S, it is denoted by  y∉Sy∉S .

Graph theory In  mathematics ,  graph theory  is the study of  graphs , which are mathematical structures used to model pairwise relations between objects . A graph in this context is made up of  vertices  (also called  nodes  or  points ) which are connected by  edges  (also called  links  or  lines ).

Graph In one restricted but very common sense of the term,  a  graph  is an  ordered pair   {G =(V,E)} comprising : V, a  set  of  vertices  ( also called   nodes  or  points ); E, a  set  of  edges  (also called  links  or  lines ), which are  unordered pairs  of vertices (that is, an edge is associated with two distinct vertices)

Directed Graph A  directed graph  or  digraph  is a graph in which edges have orientations . In one restricted but very common sense of the term ,  a  directed graph  is an ordered pair   { G=(V,E)} comprising : V, a  set  of  vertices  (also called  nodes  or  points ); E, a  set  of  edges  (also called  directed edges ,  directed links ,  directed lines ,  arrows  or  arcs ) which are  ordered pairs  of vertices (that is, an edge is associated with two distinct vertices).

Example:

 Let us consider, a Graph is G = (V, E) where V = {a, b, c, d} and E = {{a, b}, {a, c}, {b, c}, {c, d}} Here V is verteces and a, b, c, d are various vertex of the graph. Here E represents edges and {a, b}, {a, c}, {b, c}, {c, d} are various edge of the graph . Example 2:

function Definition:A function from set A to a set B is a binary relation f from A to B with a property that,for every a€A,there is exactly one b€B such that ( a,b )€f. f:A→B

Eg:A ={1,2,3,4},B={ x,y,z } f={(1,x),(2,y),(3,x),(1,z)} 1 2 3 X y

A={ a,b,c,d },B={1,2,3,4,5} 1)f={(a,1),(a,2),(b,2),(c,3),(c,4),(d,3)} Not a function a and c is repeated 2)f={(a,5),(b,3),(c,5),(d,4)} It is a function

Note:Function is also written as map. f:A→B f(1)=image of 1=x f(2)=image of 2=y f(3)=image of 3=x ( a,b )€f ↔ b=f(a) 1 2 3 X y

Domain,Codomain,Range Let f be a function from A to B f:A→B 1)Domain of f wriiten as domf is the set A. 2)Target of f/ Codomain is the set B 3)Range=collection of images {b1,b2,b3}=Range of f

Let A={1,2,3,4} and B={ a,b,c,d } and let f={(1,a),(2,a),(3,d),(4,c)} Sol: 1 2 3 4 a B C d

If f is a function from A to B,then 1)Set A is called the domain of f denoted by domf domf ={1,2,3,4} 2)Set B is called the co-domain of f. co-domain f={ a,b,c,d }

3)If ( a,b )€ f,then f(a)= b,So b is called the image of a and a is called pre-image of b. 4)The set consisting of all the images of the elements of A under the function f is called the range of f. It is denoted by f(A) F(A)={ a,c,d }

Let A={4,5,6,7,8,9} and B={ a,b,c,d,e,f } and Let={(4,a),(5,b),(6,c),(8,e),(9,a)(7,c)}

Injective Function One-one (Injective) functions are f:X→Y is one-one if images of distinct elements of X are distinct under f. A B C 1 2 3 4

A B C 1 2 3 4

Surjective Function Onto( Surjective ) function f:X→Y is onto if every element of Y is image of some element of X. a b c 1 2 3

a b c 1 2 3 4

Bijective Function A function is called bijective when it is both one-one and onto.

Injection 1)f(x)=3x-2 f(x 1 )=f(x 2 ) 3x 1 - 2 =3x 2 - 2 3 x 1 = 3 x 2 X 1 =X 2 F(x) is one to one function

2)f(x)=5x+2 3)x 2 +3 Let A={1,2,3} and B={ a,b,c,d }.In each case state whether the given function is injective or not? A)f={(1,a),(2,d),(3,b)} B)g={(1,a),(2,a),(3,d)} C)h={(1,a),(1,b),(2,d),(3,c)} D)A={(1,a),(2,b)}

Surjective Let f:R->R is defined by f(x)=2x+1 f(x)=2x+1 -> f(x)=y Y=2x+1 =2y- 2 + 2 /2 2x=y-1 = 2 y/ 2 X=y-1/2 =y y -1 2[y-1/2]+1 f(x)=y 2 2(y-1)+2/2

Let f:R->R is defined by f(x)=x+1/x-1? F(x)= y,f (x)=x+1/x-1 x=y+1/y-1 Y=x+1/x-1 f(x)=( y+1/y-1)+1 y(x-1)=x+1 (y+1/y-1)-1 yx -y=x+1 (y+1+y-1/ y-1 )/(y+1-y+1/ y-1 ) yx -x=y+1 2y+ 1 - 1 /2+ y - y x(y-1)=y+1 2 y/ 2 =y

f(x)=3x+4 F(x )=x 2 -4/x-2

f(x)=3x+4 One-one f(x 1 )=f(x 2 ) 3x 1 + 4 =3x 2 + 4 3 x 1 = 3 x 2 X 1 =x 2 F(x) is one-one

Onto f(x)=y 3y- 12 + 12 Y=3x+4 3 3x=y-4 3 y/ 3 =y x=y-4/3 f(x)=y 3(y-4/3)+4 f(x) is onto 3(y-4)+(4*3)/3

Relations A relation in mathematics defines the relationship between two different sets of information. If two sets are considered, the relation between them will be established if there is a connection between the elements of two or more non-empty sets. In the morning assembly at schools, students are supposed to stand in a queue in ascending order of the heights of all the students. This defines an ordered relation between the students and their heights.

Therefore, we can say, ‘A set of ordered pairs is defined as a relation.’

This mapping depicts a relation from set A into set B. A relation from A to B is a subset of A x B. The ordered pairs are (1,c),(2,n),(5,a),(7,n). For defining a relation, we use the notation where, set {1, 2, 5, 7} represents the domain. set {a, c, n} represents the range. Sets and Relations Sets and relation are interconnected with each other. The relation defines the relation between two given sets. If there are two sets available, then to check if there is any connection between the two sets, we use relations. For example, An empty relation denotes none of the elements in the two sets is same.

Relations in Mathematics In Maths , the relation is the relationship between two or more set of values. Suppose, x and y are two sets of ordered pairs. And set x has relation with set y, then the values of set x are called domain whereas the values of set y are called range. Example: For ordered pairs={(1,2),(-3,4),(5,6),(-7,8),(9,2)} The domain is = {-7,-3,1,5,9} And range is = {2,4,6,8} Types of Relations There are 8 main types of relations which include: Empty Relation Universal Relation Identity Relation Inverse Relation Reflexive Relation Symmetric Relation Transitive Relation Equivalence Relation

Types of Relation: Empty Relation:  A relation R on a set A is called Empty if the set A is empty set. Full Relation:  A binary relation R on a set A and B is called full if AXB. Reflexive Relation:  A relation R on a set A is called reflexive if ( a,a ) € R holds for every element a € A .i.e. if set A = { a,b } then R = {( a,a ), ( b,b )} is reflexive relation. Irreflexive relation :  A relation R on a set A is called reflexive if no ( a,a ) € R holds for every element a € A.i.e . if set A = { a,b } then R = {( a,b ), ( b,a )} is irreflexive relation. Symmetric Relation:  A relation R on a set A is called symmetric if ( b,a ) € R holds when ( a,b ) € R.i.e . The relation R={(4,5),(5,4),(6,5),(5,6)} on set A={4,5,6} is symmetric. AntiSymmetric Relation:  A relation R on a set A is called antisymmetric if ( a,b )€ R and ( b,a ) € R then a = b is called antisymmetric.i.e . The relation R = {( a,b )→ R|a ≤ b} is anti-symmetric since a ≤ b and b ≤ a implies a = b. Transitive Relation:  A relation R on a set A is called transitive if ( a,b ) € R and ( b,c ) € R then ( a,c ) € R for all a,b,c € A.i.e . Relation R={(1,2),(2,3),(1,3)} on set A={1,2,3} is transitive. Equivalence Relation:  A relation is an Equivalence Relation if it is reflexive, symmetric, and transitive. i.e. relation R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1)} on set A={1,2,3} is equivalence relation as it is reflexive, symmetric, and transitive. Asymmetric relation:  Asymmetric relation is opposite of symmetric relation. A relation R on a set A is called asymmetric if no ( b,a ) € R when ( a,b ) € R.

Types of Relations The  Empty Relation  between sets X and Y, or on E, is the empty set  ∅ ∅ The  Full Relation  between sets X and Y is the set  X × Y X×Y The  Identity Relation  on set X is the set  {( x , x )| x ∈ X } {( x,x )| x∈X } The Inverse Relation R' of a relation R is defined as −  R ′={( b , a )|( a , b )∈ R } R′={( b,a )|( a,b )∈R} Example  − If  R ={(1,2),(2,3)} R={(1,2),(2,3)} then  R ′ R′ will be  {(2,1),(3,2)} {(2,1),(3,2)} A relation R on set A is called  Reflexive  if  ∀ a ∈ A ∀a∈A  is related to a ( aRa holds) Example  − The relation  R ={( a , a ),( b , b )} R={( a,a ),( b,b )} on set  X ={ a , b } X={ a,b } is reflexive. A relation R on set A is called  Irreflexive  if no  a ∈ A a∈A  is related to a ( aRa does not hold). Example  − The relation  R ={( a , b ),( b , a )} R={( a,b ),( b,a )} on set  X ={ a , b } X={ a,b } is irreflexive . A relation R on set A is called  Symmetric  if  xRy xRy  implies  yRx yRx ,  ∀ x ∈ A ∀x∈A  and  ∀ y ∈ A ∀y∈A . Example  − The relation  R ={(1,2),(2,1),(3,2),(2,3)} R={(1,2),(2,1),(3,2),(2,3)} on set  A ={1,2,3} A={1,2,3} is symmetric. A relation R on set A is called  Anti-Symmetric  if  xRy xRy  and  yRx yRx  implies  x = y ∀ x ∈ A x = y∀x∈A  and  ∀ y ∈ A ∀y∈A . Example  − The relation  R ={( x , y )→ N | x ≤ y } R={( x,y )→ N|x≤y } is anti-symmetric since  x ≤ y x≤y  and  y ≤ x y≤x  implies  x = y x =y. A relation R on set A is called  Transitive  if  xRy xRy  and  yRz yRz  implies  xRz ,∀ x , y , z ∈ A xRz,∀x,y,z∈A . Example  − The relation  R ={(1,2),(2,3),(1,3)} R={(1,2),(2,3),(1,3)} on set  A ={1,2,3} A={1,2,3} is transitive.

A relation is an  Equivalence Relation  if it is reflexive, symmetric, and transitive. Example  − The relation R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1)}R={(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1)} on set A={1,2,3}A={1,2,3} is an equivalence relation since it is reflexive, symmetric, and transitive.

Partial Order Set Partially Ordered Sets Consider a relation R on a set S satisfying the following properties: R is reflexive, i.e., xRx for every x ∈ S. R is antisymmetric , i.e., if xRy and yRx , then x = y. R is transitive, i.e., xRy and yRz , then xRz . Then R is called a partial order relation, and the set S together with partial order is called a partially order set or POSET and is denoted by (S, ≤).

AntiSymmetric Relation Antisymmetric Relation Definition In  set theory , the relation R is said to be antisymmetric on a set A, if xRy and yRx hold when x = y. Or it can be defined as, relation R is antisymmetric if either ( x,y )∉R or ( y,x )∉R whenever x ≠ y. A relation R is not antisymmetric if there exist x,y∈A such that ( x,y ) ∈ R and ( y,a ) ∈ R but x ≠ y. Note:  If a relation is not symmetric that does not mean it is antisymmetric .

Q.1: Which of these are antisymmetric ? ( i ) R = {(1,1),(1,2),(2,1),(2,2),(3,4),(4,1),(4,4)} (ii) R = {(1,1),(1,3),(3,1)} (iii) R = {(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,1),(4,4)} Solution: ( i ) R is not antisymmetric here because of (1,2) ∈ R and (2,1) ∈ R, but 1 ≠ 2. (ii) R is not antisymmetric here because of (1,3) ∈ R and (3,1) ∈ R, but 1 ≠ 3. (iii) R is not antisymmetric here because of (1,2) ∈ R and (2,1) ∈ R, but 1 ≠ 2 and also (1,4) ∈ R and (4,1) ∈ R but 1 ≠ 4.

Reflexive Relation Example : Let A  =  {1, 2, 3} and R be a relation defined on  set A as R  = {(1, 1), (2, 2), (3, 3)} Verify R is reflexive.  Solution :  In the set A, we find three elements. They are 1, 2 and 3. When we look at the ordered pairs of R, we find the following associations.  (1, 1) -----> 1 is related to 1 (2, 2) -----> 2 is related to 2 (3, 3) -----> 3 is related to 3 In R, it is clear that every element of A is related to itself.  So, R is reflexive relation. 

Practice Problems Problem 1 :  Let A  =  {2, 3, 7}, R be a relation defined on set as R  =  {(2, 2), (3, 3), (2, 3) (3, 7)} Determine whether R is reflexive relation.

Solution :  If the relation on set A is reflexive, every element of A has to be related to itself.  In the set A, we find three elements. They are 1, 2 and 3. So, we must have 2 has to be related to itself ------> (2, 2) 3 has to be related to itself ------> (3, 3) 7 has to be related to itself ------> (7, 7) But, the ordered pair (7,7) is not in R.  So, R is not reflexive. 

Problem 2 : Let A  =  {1, 2, 3} and R be a relation defined on set A as R  = {(1, 1), (2, 2), (3, 3), (1, 2)} Determine whether R is reflexive relation.

Solution :  If the relation on set A is reflexive, every element of A has to be related to itself.  In the set A, we find three elements. They are 1, 2 and 3. So, we must have 1 has to be related to itself ------> (1, 1) 2 has to be related to itself ------> (2, 2) 3 has to be related to itself ------> (3, 3) We have all the three ordered pairs (1, 1), (2, 2) and (3, 3) in R.   So, R is reflexive. 

Problem 3 : Let A  =  {1, 2, 3} and R be a relation defined on  set A as R  = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} Determine whether R is reflexive relation.

Solution :  If the relation on set A is reflexive, every element of A has to be related to itself.  In the set A, we find three elements. They are 1, 2 and 3. So, we must have 1 has to be related to itself ------> (1, 1) 2 has to be related to itself ------> (2, 2) 3 has to be related to itself ------> (3, 3) We have all the three ordered pairs (1, 1), (2, 2) and (3, 3) in R.   So, R is reflexive. 

Symmetric Relation Example : Let A be the set of two male children in a family and R be a relation defined on set A as R  =  "is brother of". Verify whether R is symmetric.  For all elements Solution : Let a, b ∈ A.   If "a" is brother of "b", then "b" has to be brother of "a".  Clearly, R  =  {(a, b), (b, a)} So, R is symmetric. 

Problem 1 : Let A  =  {1, 2, 3} and R be a relation defined on set A as R  = {(1, 1), (2, 2), (3, 3), (1, 2)} Verify R is symmetric. 

Solution :  To verify whether R is symmetric, we have to check the condition given below for each ordered pair in R. That is,  (a, b) -----> (b, a) Let's check the above condition for each ordered pair in R.  From the table above, if R is symmetric, for the ordered pair (1, 2), we must have (2, 1) in R.  But, we don't have (2, 1) in R.  So, R is not symmetric. 

Problem 2 : Let A  =  {1, 2, 3} and R be a relation defined on set A as R  = {(1, 1), (2, 2), (1, 2), (2, 1)} Verify R is symmetric. 

To verify whether R is transitive, we have to check the condition given below for each ordered pair in R. That is,  (a, b) -----> (b, a) Let's check the above condition for each ordered pair in R.  From the table above, it is clear that R is symmetric.

Problem 3 : Let A  =  {1, 2, 3} and R be a relation defined on set A as R  = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} Verify R is symmetric. 

Problem 3 : Let A  =  {1, 2, 3} and R be a relation defined on set A as R  = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} Verify R is symmetric. 

Solution :  To verify whether R is transitive, we have to check the condition given below for each ordered pair in R. That is,  (a, b) -----> (b, a) Let's check the above condition for each ordered pair in R. 

Transitive Relation Example : Let A  =  { 1, 2, 3 } and R be a relation defined on  set A as "is less than" and R  = {(1, 2), (2, 3), (1, 3)} Verify R is transitive. 

Solution :  From the given set A, let a  =  1 b  =  2 c  =  3 Then, we have (a, b)  =  (1, 2) -----> 1 is less than 2  (b, c)  =  (2, 3) -----> 2 is less than 3  (a, c)  =  (1, 3) -----> 1 is less than 3  That is, if 1 is less than 2 and 2 is less than 3, then 1 is less than 3.  More clearly,  1R2, 2R3 -----> 1R3 Clearly, the above points prove that R is transitive. 

Problem 1 : Let A  =  {1, 2, 3} and R be a relation defined on set A as R  = {(1, 1), (2, 2), (3, 3), (1, 2)} Verify R is transitive. 

Solution :  To verify whether R is transitive, we have to check the condition given below for each ordered pair in R. That is,  (a, b),  (b, c) -----> (a, c) Let's check the above condition for each ordered pair in R. 

Problem 2 : Let A  =  {1, 2, 3} and R be a relation defined on set A as R  = {(1, 1), (2, 2), (1, 2), (2, 1)} Verify R is transitive. 

Solution :  To verify whether R is transitive, we have to check the condition given below for each ordered pair in R. That is,  (a, b),  (b, c) -----> (a, c) Let's check the above condition for each ordered pair in R.  ( a,b ) ( b,c )->( a,c ) (1,1) (1,2)->(1,2) (2,2) (2,1)->(2,1) (1,2) (2,2)->(1,2) (1,2) (2,1)->(1,1) (2,1) (1,1)->(2,1) (2,1) (1,2)->(2,2) From the table above, it is clear that R is transitive. 

Problem 3 : Let A  =  {1, 2, 3} and R be a relation defined on set A as R  = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)} Verify R is transitive. 

Solution :  To verify whether R is transitive, we have to check the condition given below for each ordered pair in R. That is,  (a, b),  (b, c) -----> (a, c) Let's check the above condition for each ordered pair in R.  ( a,b ) ( b,c )->( a,c ) (1,1) (1,2)->(1,2) (2,2) (2,1)->(2,1) (2,2) (2,3)->(2,3) (3,3) (3,2)->(3,2) (1,2) (2,2)->(1,2) (1,2) (2,1)->(1,1) (1,2) (2,3)->(1,3) In the table above, for the ordered pair (1, 2), we have both (a, b) and (b, c). But, we don't find (a, c).  That is, we have the ordered pairs (1, 2) and (2, 3) in R. But, we don't have the ordered pair  (1, 3)  in R.  So, we stop the process and conclude that R is not transitive. 

Equivalence Relation Example : Let A  =  {1, 2, 3,4} and R be a relation defined on set A as R  = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1),(4,4),(1,4)} Verify R is equivalence.  Solution :  We have to check whether the three relations reflexive, symmetric and transitive hold in R. Reflexive :  In the set A, we find three elements. They are 1, 2 and 3. When we look at the ordered pairs of R, we find the following associations.  (1, 1) -----> 1 is related to 1 (2, 2) -----> 2 is related to 2 (3, 3) -----> 3 is related to 3 In R, it is clear that every element of A is related to itself.  So, R is reflexive relation. 

Symmetric : To verify whether R is transitive, we have to check the condition given below for each ordered pair in R. That is,  (a, b) -----> (b, a) Let's check the above condition for each ordered pair in R.  From the table above, it is clear that R is symmetric.

Transitive : To verify whether R is transitive, we have to check the condition given below for each ordered pair in R. That is,  (a, b),  (b, c) -----> (a, c) Let's check the above condition for each ordered pair in R. 

Partial Order set A={1,2,3} R={} R={(1,1),(2,2),(3,3)} R={(1,1),(2,2),(3,3),(1,2),(2,1)} R={(1,1),(2,2),(3,3),(1,3),(2,3)} R={(1,1),(1,2),(2,3),(1,3)}

Q)A={1,2,3,4} R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3), (3,4),(4,4)} Sol:R is reflexive (1,1),(2,2),(3,3),(4,4)€R R is antisymmetric :(1,2)€R (2,1) € R It is not symmetric R is transitive :(1,3)€R (3,4)€R (1,4)€R It is a POSET

A={1,2,3,4,5} R={(1,1),(2,2),(3,3),(4,4)} R={(1,1),(2,2),(3,3),(4,4),(5,5),(2,1),(2,3),(1,2)} R={(1,1),(2,2),(3,3),(4,4),(5,5),(1,2)}

Hasse Diagram Used to represent POSET. Steps: 1)Create vertex for element.(upward direction). 2)If aRb then draw edge from a to b. 3)Remove Self Loop and transitive edge.

Construct The Hasse diagram for ({1,2,3,4},<=). Sol: 4. 3. 2. 1. Raw hasse diagram

Draw the Hasse Diagram representing the partial ordering {( a,b ) a divides b} on {1,2,3,4,6,8,12}

A={2,3,6,12,18,36} {a divides b} A={1,2,3,4,6,8,9,12,18,24} R={(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)}

Terminilogy of POSET Maximal Element:an element is not related to any other element.(upward direction) Minimal Element:a is minimal in (S,<=) if there is no b€S s.t b≤a .(Bottom elements of hasse diagram). Greatest Element:a is the greatest element of (S,<=) if b≤a for all b€S.It must be unique. Least Element:a is the least element of (S,<=) if a≤b for all b€S.It must be unique.

Example Which element of POSET are ({2,4,5,10,12,20,25},/) are maximal,minimal,greatest,least element. Maximal-12,20,25 Minimal-2,5 Greatest-No unique element Least-No unique element

Which element of POSET are ({2,3,4,8,10,12,20},/) are maximal,minimal,greatest,least element.

A b c a b e c d d

Recurrence Relations A sequence can be defined by giving a general formula for its nth term or by writing of its terms. An alternative approach is to write the sequence by finding a relationship among its terms such a relationship is called a recurrence relation.

Example S={5,8,11,14,17________} Initial condition:s1=5 Sn =S n-1 +3 S2=S1+3 =5+3 S2=8 S3=S2+3 =8+3 S3=11

Find the first four terms of each of the following recurrence relation. a k =4ka k-1 +3ka k-1 +2k+k for all integer k>=2,a 1 =1 a 1 =1 a 2 =4ka k-1 +3ka k-1 +2k+k = 4*2*1+3*2*1+2*2+2 =8+6+4+2 = 20 a 3 =4ka k-1 +3ka k-1 +2k+k = 4*3*20+3*3*20+2*3+3 = 240+180+6+3=429 a 4 =4ka k-1 +3ka k-1 +2k+k = 4*4*429+3*4*429+2*4+4 = 6864+5148+8+4 =12024 { 1,20,429,12024_______}
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