MTH 180 Concepts#

Additional Resources

Disclaimer

The following sets of videos have been vetted for the general knowledge of mathematical subject matter applicable to CSC 212 but not directly focused on. This is not an exhaustive list, but a guide to helpful instruction. The variety in selection is merely to help fit the multitude of learning curves for students within the course.

Sequences#

Sequence#

a function whose domain is the set of positive integers
ex: {1,2,3,4,}

The general, or nth term, is denoted as an

odd integers

1

3

5

7

9

n

perfect squares

1

4

9

16

25

n

a1

a2

a3

a4

a5

an

Try it

Write out the first five (5) terms of the sequence…

an=2n+3
a1=2(1)+3=5a2=2(2)+3=7a3=2(3)+3=9a4=2(4)+3=11a5=2(5)+3=13
first five (5) terms of the sequence 
{5,7,9,11,13}
an=(1)n62nn+1
a1=(1)162(1)(1)+1=42=422=22
a2=(1)262(2)(2)+1=23=233
a3=(1)362(3)(3)+1=04=0
a4=(1)462(4)(4)+1=25=255
a5=(1)562(5)(5)+1=46=263
first five (5) terms of the sequence 
{22,233,0,255,263}

Try it

Write the nth term of the sequence {an} suggested by the pattern

23,49,827,1681,

Find the pattern…

2131,2232,2333,2434

Therefore,

an=2n3n=(23)n
2,4,6,8,10,

Find the pattern…

2

4

6

8

10

n

a1

a2

a3

a4

a5

an

Check : an=(1)n2n
a1=(1)12(1)=2
Check : an=(1)n+12n
a1=(1)n+12(1)=122=2

Therefore,

an=(1)n+12n
Recursive relation#

a sequence for which the given is the first term together with a rule for obtaining any an+1 from the preceding term an for n1

Try it

Find the first five (5) terms of the recursively defined infinite sequence

a1=2,     an+1=nan
a1=2
a2

an+1=a2a2=1a1=12an+1=a3
a3

a3=2a2=212=4
a4

a4=3a3=34
a5

a5=4a4=4(34)=163
first five recursive terms

12,4,34,163     
a1=1,     a2=1,     an=an+1+nan1
a1=1a2=1
a3

a3=a32+(3a31)=a1+(3a2)=1+3(1)=2
a4

a4=a42+(4a41)=a2+(4a3)=1+4(1)=9
a3

a5=a52+(5a51)=a3+(5a4)=2+5(9)=47
first five recursive terms

1,1,4,9,47

Series#

Series#

the sum of a number of terms in a sequence

can be finite

1+3+5+7
can be infinite

1+3+5+7+
1+12+14+18+
Summation / Sigma Notation
https://knowledgeuniverseonline.com/wp-content/uploads/2022/03/sigma-notation.png
kindex ofsummation

n = stopping point
= sigma or “sum of”
k = starting point
ak = formula for each term

Example


k=1nak=a1+a2+a3+a4++an1+an

Try it

Write out each sum (expand and simplify)

k=163k+1
k=163k+1=31+1+32+1+33+1+34+1+35+1+36+1=32+1+34+35+12+37=669140

Try it

Express each sum using summation notation

1+3+5+7++[2(12)1]
1+3+5+7++[2(12)1]a1+a2+a3+a4++a12
Check : ak=2k1
a1=211=1a2=221=3a3=231=5a4=241=7
Therefore
k=1122k1

Try it

Find the sum for each sequence

k=124k
=(1)+(2)+(3)+(4)++(24)=(1+2+3+4++24)=24(24+1)2=12(25)=300

Formula

k=1nk=1+2+3++n=n(n+1)2
k=1374282.1
(2.1+2.1++2.1)=(2.1)(292)=613.2a137a428   428137   291 iterations

Formula

k=mnc=(nm+1)ck=1374282.1=(428137+1)2.1

c = constant

k=014k24
(024)+(124)+(224)++(1424)=955

Formula

k=1n(akbk)=k=1nakk=1nbkk=014(k24)=k=014k2k=0144

Breakout 1

k=1nk2=12+22+32++n2=n(n+1)(2n+1)6=14(14+1)(2(14)+1)6=1015

Breakout 2

k=1n4=4(15)=60

Outcome

k=014(k24)=k=014k2k=0144=101560=955