MAS162: Foundations of Discrete Mathematics Assignment 1 Solutions

This document provides a comprehensive solution to Assignment 1 for the MAS162: Foundations of Discrete Mathematics course. The solution includes answers to several mathematical problems, such as finding recursive definitions for given sequences. The assignment also explores saving schemes, with detailed calculations and MATLAB code implementations to model and analyze different scenarios. Furthermore, the document delves into the concept of mathematical induction, providing step-by-step solutions and explanations. The solutions are presented clearly, demonstrating the application of mathematical principles and the use of computational tools to solve problems related to sequences, series, and mathematical proofs. The assignment covers topics like recursive definitions, saving schemes, and MATLAB code.

MATHS
Student Name –
Student ID –

Solution 1 :
a) Simple Recursive Definition of the sequence :
tn = 7 / 5 –n+1
tn-1 = 7 / 5 –(n-1)+1
Dividing the above expressions:
tn = 5 tn-1
b) Simple Recursive Definition of the sequence :
tn = 7 + 3 (n-1)
tn-1 = 7 + 3 (n-1-1)
Subtracting both the equations:
tn = tn-1 + 3
c) Simple Recursive Definition of the sequence :
tn = (n-1)! / 7 –(n-1)
tn+1 = n! / 7 –(n)
Dividing the above expressions:
tn+1 = 7n tn

d) Formula for nth term :
. t2 = 8 / ∏
Tn = - ∏ + tn-1
T3 = - ∏ + 8 / ∏
T4 = - 2 ∏ + 8 / ∏
Tn = - ( n-2 ) ∏ + 8 / ∏
Solution 2 :
Saving Schemes :
a) A1 = 2300 x 1.03
A2 = (A1 + 150) x 1.03
A3 = (A2 + 150) x 1.03
An+1 = (An + 150) x 1.03
b) B1 = ( 2100 x 1.04 ) + 130
B2 = ( B1 x 1.04 ) + 130
Bn+1 = ( Bn x 1.04 ) + 130

c) MATLAB CODE:
A(1) = 2300*1.03;
B(1) = (2100*1.04)+130;
for i=1:10
A(i+1) = (A(i)+150)*1.03;
B(i+1) = (B(i)*1.04)+130;
end
display('n An Bn');
for i=1:10
display (i);
display (A(i));
display (B(i));
end
d) MATLAB CODE :
A(1) = 2300*1.03;
B(1) = (2100*1.04)+130;
for i=1:10
A(i+1) = (A(i)+150)*1.03;
B(i+1) = (B(i)*1.04)+130;

end
i=1:10;
subplot(2,1,1);
plot (i,A(i));
xlabel('n');
ylabel('An');
subplot(2,1,2);
plot (i,B(i));
xlabel('n');
ylabel('Bn');
e)
MATLAB CODE:
A(1) = 2300*1.03;
B(1) = (2100*1.04)+130;
for i=1:10
A(i+1) = (A(i)+150)*1.03;
B(i+1) = (B(i)*1.04)+130;
end
d1 = 2300 + (150* 9);
d2 = 2100 + (130 * 10);
display('Total Deposit for Case1 :');

display (d1);
display('Total Deposit for Case2 :');
display (d2);
i1 =
0.03*(2300+(150*9)+A(1)+A(2)+A(3)+A(4)+A(5)+A(6)+A(7)+A(8)+A(9));
i2 = 0.04*(2100+ B(1)+B(2)+B(3)+B(4)+B(5)+B(6)+B(7)+B(8)+B(9));
display('Total Interest for Case1 :');
display (i1);
display('Total Interest for Case2 :');
display (i2);

Figure 1
Solution 3)
a)
. t1 = β a1
. . t2 = β a2
. . t3 = β a3
. . tn-1 = β an-1

. tn = β an
. tn+1 = β an+1
b)
. s1 = t1 = β a1
. s2 =t1 + t2 + = β a1 + β a2
. s3 =t1 + t2 + t3 + = β a1 + β a2 + β a3
. sn-1 =t1 + t2 +.... tn-1 + = β a1 + β a2 + ........... β an-1
. sn =t1 + t2 +...... tn = β a1 + β a2 + ........... β an
. sn+1 =t1 + t2 +...... tn+1 = β a1 + β a2 + ........... β an+1
c)
. sn+1 =t1 + t2 +...... tn+1 = β a1 + β a2 + ........... β an+1
Multiplying given equation by ‘a’ :
. a sn+1 = β a2 + β a3 + ........... β an+2
Subtracting both the expressions:
a sn+1 - sn+1 = β an+2 - β a1
sn+1 = β a ( 1 - an+1 ) / (1-a)
d)
By using general term :

MATLAB Code:
a= 3;
b= 2;
for i=0:7
s(i+1) = (b*a)*(1-(a^(i+1)))/ (1-a);
display (s(i+1));
end
By using iteration :
MATLAB Code:
a= 3;
b= 2;
s(1) = 2 * 3;
display(s(1));
for i=1:7
s(i+1) = s(i) + (b * (a^(i+1)));
display (s(i+1));
end
Solution 4)

a)
Figure 2
The values are :
T0 = 1, t1 = 2, t2 = 4 , t3= 7, t4 = 11

The cut lines are placed in such a manner that they intersect other lines at
maximum number of points.
b)
It is found that ‘n’ new pieces are formed if n-th line is drawn.
Recursive definition :
. t0 = 1
. t1 = 1+1 = 2
. t2 = 2+2 = 4
. t3 = 4 + 3 = 7
. t4 = 7+4 = 11
Hence,
. tn = tn-1 + n
c)
tn = (n2 + n + 2)/2
tn-1 = ((n-1)2 + n-1 + 2)/2
. tn-1 = tn – n
Hence,

. tn = tn-1 + n
This is the same relation as obtained in part ‘b’.
Hence, the function given is valid.