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STARTS WITH THE NAME OF 
ALLAH 
the most beneficent & Merciful
PRESENTAION 
AQEEL RAFIQUE 01 
HAMZA MAQSOOD 12 
REYAN IQBAL 28 
UMAIR HAIDRI 17 
SHOAIB ASHRAF 44 
RAJA AMIR 43 
PRESENTED BY :
Binomial Theorem
Binomial : 
In algebra (a+b) is called binomial. 
Binomial theorem provides an expression for the power of 
binomial of n.
Binomial Theorem 
The expansion of binomial theorem is. 
푎 + 푏 푛 = 
푛 
푘=0 
푛 
푘 
푎푛−푘 푏푘 
푎 + 푏 푛 =an+ 
푛 
1 푎푛−1푏1 + 푛 
2 푎푛−2푏2 + ⋯ + 푛 
푛−1 푎1푏푛−1+bn
Binomial Theorem 
Need of Binomial Theorem 
As we know 
(a+b)0 =1 
(a+b)1=(a+b) 
(a+b)2 =a2+2ab+b2 
. 
. 
(a+b)n=?
Binomial Theorem 
푎 + 푏 0 = 
0 
푎 + 푏 푛 = 
푘=0 
푛 
푘=0 
Proof by Algebraic method 
0 
푘 푎0−0푏0 = 1 
Let (1) is true for n=m 
푎 + 푏 푚 = 
푛 
푘 푎푛−푘푏푘 (1) 
푚 
푘=0 
푚 
푘 
푎푚−푘푏푘 
If true for n=m then also true for n=m+1
Binomial Theorem 
푎 + 푏 푚+1 = 
푚+1 
푘=0 
푚+1 
푘 푎푚+1−푘푏푘 
R.H.S 
푎 + 푏 푚+1 = 푎 + 푏 푎 + 푏 푚 
=(a + b) 
푚 
푘=0 
푚 
푘 푎푚−푘푏푘 
=a . 
푚 
푘=0 
푚 
푘 푎푚−푘푏푘+b . 
푚 
푘=0 
푚 
푘 푎푚−푘푏푘 
= 
푚 
푘=0 
푚 
푘 푎푚+1−푘푏푘+ 
푚 
푘=0 
푚 
푘 푎푚−푘푏푘+1 
Replacing variables: 
Let j=k+1; k=j-1 when k=0, j=1 
when k=m, j=m+1
Binomial Theorem 
푎 + 푏 푚+1 = 
푚 
푘=0 
푚 
푘 푎푚+1−푘푏푘 + 
푚 
푘=0 
푚 
푘 푎푚−푘푏푘+1 
Taking second summation on the right hand side. 
= 
푚 
푗−1 푎푚−(푗−1)푏푗−1+1 
= 
푚 
푗−1 푎푚+1−푗)푏푗 
= 
푚+1 
푗=1 
푚 
푗−1 푎푚−푗+1푏푗 
= 
푚+1 
푘=1 
푚 
푘−1 푎푚+1−푘푏푘 
푎 + 푏 푚+1 = 
푚 
푘=0 
푚 
푘 푎푚+1−푘푏푘 + 
푚+1 
푘=1 
푚 
푘−1 푎푚+1−푘푏푘 
=am+1+ 
푚+1 
푘=1 
푚 
푘 + 푚 
푘−1 푎푚+1−푘푏푘+bm+1
Binomial Theorem 
푎 + 푏 푚+1 = am+1+ 
푚+1 
푘=1 
푚+1 
푘 푎푚+1−푘푏푘 +bm+1 by Pascal’s formula 
푎 + 푏 푚+1 = 
푚+1 
푘=0 
푚+1 
푘 푎푚+1−푘푏푘 
Hence proved. 
This shows that if any number in the power of binomial is given we can easily 
find its expansion.
Counting elements 
in one dimensional 
array.
Counting elements in one dimensional 
array. 
Let A[1],A[2],A[3]……………….A[n] is a one dimensional array. Where 
n a positive integer. 
To find the number of element in one dimensional array by using the 
theorem.
Counting elements in one dimensional 
array. 
We use theorem of find the no of elements in a list. 
i-e:- n-m+1 
where n is the last term of the list and m is the first term of 
the list.
Counting elements in one dimensional 
array. 
Example: 
suppose the elements in 1 dimensional array are; 
A[2]=2; 
A[3]=3; 
A[4]=5; 
. 
. 
. 
A[10]=7
Counting elements in one dimensional 
array. 
By Applying theorem we get 
Apply theorem on index; 
Where n=10, m=2; 
The number of elements in the given array are: 
n-m+1=10-2+1 
=9 
Elements = 9
Recursion
Recursion 
Recursively Defined Sequence 
Method of defining a sequence: 
 Informal ways 
 Explicit formula 
 Recursion
Recursion 
Recursively Defined Sequence 
 Informal way: 
In informal ways a sequence is given we extract or 
generate the pattern of the sequence and generate the next 
term. 
 Disadvantages: 
• Misunderstand of the sequence cause the error. 
For example: 
if the sequence 3,5,7……. Is given if some one 
guess it prime number place 9 if someone understand it 
prime number he put 11. so this cause the 
misunderstanding.
Recursion 
Recursively Defined Sequence 
 Explicit formula: 
In explicit formula we make a formula for the nth 
term of the sequence. 
For example: 2,4,6………. 
Explicit formula for above sequence is 2k, where k>0 
 Advantages: 
• Each term is uniquely determine. 
 Disadvantage: 
• Difficult to make the explicit formula if such 
sequence is given which is difficult to analyze.
Recursion 
Recursively Defined Sequence 
Recursion: 
In recursion two equations are given. 
• Recurrence relation: 
It is the formula to find the sequence. 
• Initials conditions: 
it is the first few values of the sequence. It is also called 
base or bottom of the recursion.
Recursion 
Recursively Defined Sequence 
 For example: 
1) bk = bk-1+ bk-2 recurrence relation 
2)b0=1, b1=3 initial conditions 
Every founded value is used to find the next term of the sequence
For example: 
A sequence c0, c1, c2, . . . recursively as follows: For all integers 
k ≥ 2, 
(1) ck = ck−1 + k.ck−2 + 1 recurrence relation 
(2) c0 = 1 and c1 = 2 initial conditions. 
Find c2,c3. 
Recursion 
Computing Terms of a Recursively Defined Sequence
Computing Terms of a Recursively Defined Sequence 
Solution: 
Recursion 
Putting k=2 
since c1 = 2 and c0 = 1 
c2 = c1 + 2c0 + 1 
= 2 + 2·1 + 1 
=5 
similarly for c3 we put k=3 and solve.
Recursion 
Sequences That Satisfy the Same Recurrence Relation 
Let a1, a2, a3, . . . and b1, b2, b3, . . . satisfy the recurrence relation 
that the kth term equals 3 times the (k − 1) term for all integers 
k ≥ 2 
(1) ak = 3ak−1 and bk = 3bk−1 
And the initial conditions are: 
a1=3, b1=1 
Find a2, a3 and b2 ,b3
Recursion 
Sequences That Satisfy the Same Recurrence Relation 
Solution: 
a2 = 3a1 = 3·3 = 9 
a3 = 3a2 = 3·9 = 27 
So the ‘a’ sequence is 3,9,27……. 
b2 = 3b1 = 3·1 = 3 
b3 = 3b2 = 3·3 = 9 
So the ‘b’ sequence is 1,3,9,………
Tower of Hanoi 
The Tower of Hanoi (also called the Tower of 
Brahma or Lucas' Tower, and sometimes 
pluralized) is a mathematical game or puzzle. It 
consists of three rods, and a number of disks of 
different sizes which can slide onto any rod. The 
puzzle starts with the disks in a neat stack in 
ascending order of size on one rod, the smallest at 
the top, thus making a conical shape.
Tower of Hanoi 
A B C 
1 
n-1 
n
Tower of Hanoi 
Objective 
The objective of the puzzle is to move the entire stack to 
another rod, obeying the following simple rules: 
• Only one disk can be moved at a time. 
• Each move consists of taking the upper disk from one of 
the stacks and placing it on top of another stack i.e. a disk 
can only be moved if it is the uppermost disk on a stack. 
• No disk may be placed on top of a smaller disk.
Recursive pattern: 
From the moves necessary to transfer one, two, and three disks, we can 
find a recursive pattern - a pattern that uses information from one step to 
find the next step - for moving n disks from post A to post C: 
First, transfer n-1 disks from post A to post B. The number of moves will 
be the same as those needed to transfer n-1 disks from post A to post 
C. Call this number M moves. 
Next, transfer disk 1 to post C [1 move]. 
Finally, transfer the remaining n-1 disks from post B to post C. [Again, the 
number of moves will be the same as those needed to transfer n-1 
disks from post A to post C, or M moves.] 
No of moves : (n-1)+1+(n-1) 
2(n-1)+1
Tower of Hanoi 
What? 
if we want to know how many moves it will take to transfer 100 disks 
from post A to post B. 
Ans: Through recursion we will first have to find the moves it takes to 
transfer 99 disks, 98 disks, and so on. 
So now we find the explicitly: 
Number of Disks Number of Moves 
1 1 
2 3 
3 7 
4 15 
5 31 
The pattern generated from this sequence is: 2n-1
Tower of Hanoi 
1,3,7,15……………. 
1+0=1 21-1=1 
1+2=3 22-1=3 
3+4=7 23-1=7 
7+8=15 24-1=15 
. . 
. . 
. . 
So the minimum number of moves required to solve a 
Tower of Hanoi puzzle is 2n - 1, where n is the number of 
disks.
Relations
Relations 
A relation R is a subset of the Cartesian product of the 
given set(s). 
Given in order pair form (x , y). 
x related to y, if and only if (x , y) is in R. 
Denoted by x R y.
Relations 
Properties of Relations: 
 Reflexive 
 Symmetric 
 Transitive
Relations 
Properties of Relations: 
Let R be a relation on set A 
 Reflexive: 
R is reflexive if and only if x R x for all x is in A. 
 Symmetric: 
R is symmetric if x R y then y R x ; x , y is in A 
 Transitive: 
R is transitive if x R y and y R x then x R z; x , y , z is in A
Relations 
Properties of Relations: 
Let A = {0, 1, 2, 3} and define relations R on A as 
follows: 
R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}, 
Is R reflexive, symmetric, transitive ? 
Solution: 
Graph of relation will be
Relations 
Properties of Relations: 
Reflexive: 
R is reflexive because each element contain loop, mean each element is 
related to itself 
Symmetric: 
R is symmetric because here an arrow move from one point to second and 
also from second to first, mean first related to second and also second 
related to first. 
Transitive: 
R is not transitive because there is no arrow moves from 3 to 1. so 1 R 0 and 
0 R 3 but 1 R 3.
Any 
Question?
Thanks 
!

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Binomial Theorem, Recursion ,Tower of Honai, relations

  • 1. STARTS WITH THE NAME OF ALLAH the most beneficent & Merciful
  • 2. PRESENTAION AQEEL RAFIQUE 01 HAMZA MAQSOOD 12 REYAN IQBAL 28 UMAIR HAIDRI 17 SHOAIB ASHRAF 44 RAJA AMIR 43 PRESENTED BY :
  • 4. Binomial : In algebra (a+b) is called binomial. Binomial theorem provides an expression for the power of binomial of n.
  • 5. Binomial Theorem The expansion of binomial theorem is. 푎 + 푏 푛 = 푛 푘=0 푛 푘 푎푛−푘 푏푘 푎 + 푏 푛 =an+ 푛 1 푎푛−1푏1 + 푛 2 푎푛−2푏2 + ⋯ + 푛 푛−1 푎1푏푛−1+bn
  • 6. Binomial Theorem Need of Binomial Theorem As we know (a+b)0 =1 (a+b)1=(a+b) (a+b)2 =a2+2ab+b2 . . (a+b)n=?
  • 7. Binomial Theorem 푎 + 푏 0 = 0 푎 + 푏 푛 = 푘=0 푛 푘=0 Proof by Algebraic method 0 푘 푎0−0푏0 = 1 Let (1) is true for n=m 푎 + 푏 푚 = 푛 푘 푎푛−푘푏푘 (1) 푚 푘=0 푚 푘 푎푚−푘푏푘 If true for n=m then also true for n=m+1
  • 8. Binomial Theorem 푎 + 푏 푚+1 = 푚+1 푘=0 푚+1 푘 푎푚+1−푘푏푘 R.H.S 푎 + 푏 푚+1 = 푎 + 푏 푎 + 푏 푚 =(a + b) 푚 푘=0 푚 푘 푎푚−푘푏푘 =a . 푚 푘=0 푚 푘 푎푚−푘푏푘+b . 푚 푘=0 푚 푘 푎푚−푘푏푘 = 푚 푘=0 푚 푘 푎푚+1−푘푏푘+ 푚 푘=0 푚 푘 푎푚−푘푏푘+1 Replacing variables: Let j=k+1; k=j-1 when k=0, j=1 when k=m, j=m+1
  • 9. Binomial Theorem 푎 + 푏 푚+1 = 푚 푘=0 푚 푘 푎푚+1−푘푏푘 + 푚 푘=0 푚 푘 푎푚−푘푏푘+1 Taking second summation on the right hand side. = 푚 푗−1 푎푚−(푗−1)푏푗−1+1 = 푚 푗−1 푎푚+1−푗)푏푗 = 푚+1 푗=1 푚 푗−1 푎푚−푗+1푏푗 = 푚+1 푘=1 푚 푘−1 푎푚+1−푘푏푘 푎 + 푏 푚+1 = 푚 푘=0 푚 푘 푎푚+1−푘푏푘 + 푚+1 푘=1 푚 푘−1 푎푚+1−푘푏푘 =am+1+ 푚+1 푘=1 푚 푘 + 푚 푘−1 푎푚+1−푘푏푘+bm+1
  • 10. Binomial Theorem 푎 + 푏 푚+1 = am+1+ 푚+1 푘=1 푚+1 푘 푎푚+1−푘푏푘 +bm+1 by Pascal’s formula 푎 + 푏 푚+1 = 푚+1 푘=0 푚+1 푘 푎푚+1−푘푏푘 Hence proved. This shows that if any number in the power of binomial is given we can easily find its expansion.
  • 11. Counting elements in one dimensional array.
  • 12. Counting elements in one dimensional array. Let A[1],A[2],A[3]……………….A[n] is a one dimensional array. Where n a positive integer. To find the number of element in one dimensional array by using the theorem.
  • 13. Counting elements in one dimensional array. We use theorem of find the no of elements in a list. i-e:- n-m+1 where n is the last term of the list and m is the first term of the list.
  • 14. Counting elements in one dimensional array. Example: suppose the elements in 1 dimensional array are; A[2]=2; A[3]=3; A[4]=5; . . . A[10]=7
  • 15. Counting elements in one dimensional array. By Applying theorem we get Apply theorem on index; Where n=10, m=2; The number of elements in the given array are: n-m+1=10-2+1 =9 Elements = 9
  • 17. Recursion Recursively Defined Sequence Method of defining a sequence:  Informal ways  Explicit formula  Recursion
  • 18. Recursion Recursively Defined Sequence  Informal way: In informal ways a sequence is given we extract or generate the pattern of the sequence and generate the next term.  Disadvantages: • Misunderstand of the sequence cause the error. For example: if the sequence 3,5,7……. Is given if some one guess it prime number place 9 if someone understand it prime number he put 11. so this cause the misunderstanding.
  • 19. Recursion Recursively Defined Sequence  Explicit formula: In explicit formula we make a formula for the nth term of the sequence. For example: 2,4,6………. Explicit formula for above sequence is 2k, where k>0  Advantages: • Each term is uniquely determine.  Disadvantage: • Difficult to make the explicit formula if such sequence is given which is difficult to analyze.
  • 20. Recursion Recursively Defined Sequence Recursion: In recursion two equations are given. • Recurrence relation: It is the formula to find the sequence. • Initials conditions: it is the first few values of the sequence. It is also called base or bottom of the recursion.
  • 21. Recursion Recursively Defined Sequence  For example: 1) bk = bk-1+ bk-2 recurrence relation 2)b0=1, b1=3 initial conditions Every founded value is used to find the next term of the sequence
  • 22. For example: A sequence c0, c1, c2, . . . recursively as follows: For all integers k ≥ 2, (1) ck = ck−1 + k.ck−2 + 1 recurrence relation (2) c0 = 1 and c1 = 2 initial conditions. Find c2,c3. Recursion Computing Terms of a Recursively Defined Sequence
  • 23. Computing Terms of a Recursively Defined Sequence Solution: Recursion Putting k=2 since c1 = 2 and c0 = 1 c2 = c1 + 2c0 + 1 = 2 + 2·1 + 1 =5 similarly for c3 we put k=3 and solve.
  • 24. Recursion Sequences That Satisfy the Same Recurrence Relation Let a1, a2, a3, . . . and b1, b2, b3, . . . satisfy the recurrence relation that the kth term equals 3 times the (k − 1) term for all integers k ≥ 2 (1) ak = 3ak−1 and bk = 3bk−1 And the initial conditions are: a1=3, b1=1 Find a2, a3 and b2 ,b3
  • 25. Recursion Sequences That Satisfy the Same Recurrence Relation Solution: a2 = 3a1 = 3·3 = 9 a3 = 3a2 = 3·9 = 27 So the ‘a’ sequence is 3,9,27……. b2 = 3b1 = 3·1 = 3 b3 = 3b2 = 3·3 = 9 So the ‘b’ sequence is 1,3,9,………
  • 26. Tower of Hanoi The Tower of Hanoi (also called the Tower of Brahma or Lucas' Tower, and sometimes pluralized) is a mathematical game or puzzle. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.
  • 27. Tower of Hanoi A B C 1 n-1 n
  • 28. Tower of Hanoi Objective The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules: • Only one disk can be moved at a time. • Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack. • No disk may be placed on top of a smaller disk.
  • 29. Recursive pattern: From the moves necessary to transfer one, two, and three disks, we can find a recursive pattern - a pattern that uses information from one step to find the next step - for moving n disks from post A to post C: First, transfer n-1 disks from post A to post B. The number of moves will be the same as those needed to transfer n-1 disks from post A to post C. Call this number M moves. Next, transfer disk 1 to post C [1 move]. Finally, transfer the remaining n-1 disks from post B to post C. [Again, the number of moves will be the same as those needed to transfer n-1 disks from post A to post C, or M moves.] No of moves : (n-1)+1+(n-1) 2(n-1)+1
  • 30. Tower of Hanoi What? if we want to know how many moves it will take to transfer 100 disks from post A to post B. Ans: Through recursion we will first have to find the moves it takes to transfer 99 disks, 98 disks, and so on. So now we find the explicitly: Number of Disks Number of Moves 1 1 2 3 3 7 4 15 5 31 The pattern generated from this sequence is: 2n-1
  • 31. Tower of Hanoi 1,3,7,15……………. 1+0=1 21-1=1 1+2=3 22-1=3 3+4=7 23-1=7 7+8=15 24-1=15 . . . . . . So the minimum number of moves required to solve a Tower of Hanoi puzzle is 2n - 1, where n is the number of disks.
  • 33. Relations A relation R is a subset of the Cartesian product of the given set(s). Given in order pair form (x , y). x related to y, if and only if (x , y) is in R. Denoted by x R y.
  • 34. Relations Properties of Relations:  Reflexive  Symmetric  Transitive
  • 35. Relations Properties of Relations: Let R be a relation on set A  Reflexive: R is reflexive if and only if x R x for all x is in A.  Symmetric: R is symmetric if x R y then y R x ; x , y is in A  Transitive: R is transitive if x R y and y R x then x R z; x , y , z is in A
  • 36. Relations Properties of Relations: Let A = {0, 1, 2, 3} and define relations R on A as follows: R = {(0, 0), (0, 1), (0, 3), (1, 0), (1, 1), (2, 2), (3, 0), (3, 3)}, Is R reflexive, symmetric, transitive ? Solution: Graph of relation will be
  • 37. Relations Properties of Relations: Reflexive: R is reflexive because each element contain loop, mean each element is related to itself Symmetric: R is symmetric because here an arrow move from one point to second and also from second to first, mean first related to second and also second related to first. Transitive: R is not transitive because there is no arrow moves from 3 to 1. so 1 R 0 and 0 R 3 but 1 R 3.