8.1 Sequences and Series
Write the first 5 terms of the sequence. (Assume n begins with 1).
a_n=((-1)^n)/(2n-1)
a_1=(-1)/1=-1
a_2=1/(3)
a_3=(-1)/5
a_4=1/7
a_5=(-1)/9
Determine whether the sequence is arithmetic. If it is, find the common difference.
5, 3, 1, -1, -3,...
Yes it is arithmetic and the common difference is
d=-2
Determine whether the sequence is Geometric or not. If it is, find the common ratio.
54, -18, 6, -2, ...
It is and the common ratio is
r=-1/3
Use pascals triangle to find the binomial coefficient.
._8C_4
Remember, the rows and columns start with 0
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
Row 6: 1 6 15 20 15 6 1
Row 7: 1 7 21 35 35 21 7 1
Row 8: 1 8 28 56 70 56 28 8 1
._8C_4=70
Slips of paper numbered 1 through 14 are placed in a hat. In how many ways can two numbers be drawn so that the sum of the numbers is 12?
Order matters
1+11, 2+10, 3+9, 4+8, 5+7, 7+5, 8+4, 9+3, 10+2, 11+1
10 ways
You randomly select a card from a 52-card deck. What is the probability that it is not a club.
n(S)=52
13 cards are clubs so 3/4 of the deck is not clubs
n(E)=(.75)*52=39
(n(E))/(n(S))=39/52
Simplify the factorial expression
(2!*5!)/(6!)
(2*5!)/(6*5!)=2/6=1/3
Write the first five terms of the arithmetic sequence
a_4=10,a_(10)=28
a_(10)=a_4+6d=28=10+6d
18=6d
d=3
a_5=13
a_4=10
a_3=7
a_2=4
a_1=1
Find the nth term of the geometric sequence
a_3=6, a_4=1
a_4=a_3*r=1=6*r
1/6=r
a_4=a_1*r^3=1=a_1*(1/6)^3=a_1/216
a_1=216
a_n=216*(1/6)^(n-1)
Find the binomial coefficient BY HAND. Then you can check with a calculator.
._12C_5
._12C_5=(12!)/((12-5)!(5)!)=(12*11*10*9*8*7!)/((7!)(5!))=(12*11*10*9*8)/(5*4*3*2*1)=95040/120=792
A college student is preparing a course schedule for the next semester. The student must select one of four mathematics courses, one of six biology courses, and one of two art courses. How many schedules are possible?
Order doesn't matter so use n choose r
Choose one out of 4 math classes (5C1)
Choose one out of 6 bio classes (6C1)
Choose one out of 2 art classes (2C1)
multiply them together
._4C_1*_6C_1*_2C_1=4*6*2=48ways
A man has five pairs of socks (no two pair are of the same color) he randomly selects two socks from a drawer. What is the probability that he gets a matching pair?
Order doesn't matter (nCr)
Total number of outcomes is
._10C_2=45=n(S)
Since there are 5 matching pairs, there are 5 ways to draw a matching pair.
5=n(E)
(n(E))/(n(S))=5/45=1/9
Find the sum
sum_(j=0)^(4)(j^2+1)
a_0=1
a_1=2
a_2=5
a_3=10
a_4=17
1+2+5+10+17=35
Find the formula for the nth term of the sequence as a function of n.
a_2=14,a_6=22
a_6=a_2+4d=22=14+4d
8=4d
d=2
a_1=a_2-d=14-2=12
a_n=12+2(n-1)=10+2n
Find the sum.
\sum_(i=1)^(15)20(0.2)^(i-1)
r=0.2
a_1=20
n=15
S_15=20((1-(0.2)^15)/(1-0.2))=20/0.8=25
Use the binomial Theorem to expand and simplify your answer.
(y-3)^3
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
a=y
b=-3
1a^3+3a^2b+3ab^2+1b^3
1(y)^3+3(y)^2(-3)+3(y)(-3)^2+1(-3)^3
y^3-9y^2+27y-27
Eight people are boarding an aircraft. Two have tickets for first class and board before those in economy class. In how many ways can the eight people board the aircraft?
Order matters so use permutations
the people in first class can board in
._2P_2=(2!)/((2-2)!)=(2!)=2" ways"
The people in economy can board in
._6P_6=(6!)/((6-6)!)=(6!)=6*5*4*3*2*1=720" ways"
Multiply those together 720*2 = 1440 different ways
A child returns a five-volume set of books to a bookshelf. The child is not able to read and so cannot distinguish one volume from another. What is the probability that the books are shelved in the correct order?
Order matters so use permutations(nPr)
Total amount of permutations is
n=5
r=5
n(S)=_5P_5=(5!)=120 ways
There is only one way that the books can be put back correctly so
n(E)=1
(n(E))/(n(S))=1/120
Use sigma notation to write the sum. Then use a calculator to find the sum.
1/2+2/3+3/4+...+9/10
\sum_(x=1)^9(x/(x+1))
After plugging in your calculator.
\sum_(x=1)^9(x/(x+1))=7.071
Find the sum of the first 100 positive multiples of 5
a_1=5
d=5
a_n=5+5(n-1)=5n
a_100=5(100)=500
S_100=(100/2)(5+500)=50(505)=25250
Find the sum of the series if possible
\sum_(k=1)^(infty)4(2/3)^(k-1)
r=2/3
|r|<1
so the series converges
a_1=4
S=4/(1-(2/3))=4/(1/3)=12
Use the binomial Theorem to expand and simplify your answer.
(a-4b)^5
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
a=a
b=-4b
1a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+1b^5
1a^5+5a^4(-4b)+10a^3(-4b)^2+10a^2(-4b)^3+5a(-4b)^4+1(-4b)^5
a^5-20a^4b + 160a^3b^2-640a^2b^3+1280ab^4-1080b^5
From a pool of 7 juniors and 11 seniors, four co-captains will be chosen for the football team. How many different combinations are possible if two juniors and two seniors are to be chosen?
Order does matter so do choice (nCr)
Juniors- choose 2 from 7
n=7, r=2 (7C2)
Seniors - choose 2 from 11
n=11, r=2 (11C2)
._7C_2*_11C_2=(7!)/((7-2)!(2!))*(11!)/((11-2)!(2!))=21*55=1155ways
You select a card from a deck of 52 cards. What is the probability that it will be a red face card but not a jack?
The amount of red face cards is 6, but there are two red jacks so the n(E)=4
(n(E))/(n(S))=4/52
A deposit of $2500 is made in an account that earns 8% interest compounded quarterly. The balance in the account after n quarters is given by
a_n=2500(1+(0.08)/(4))^n, n=1, 2, 3, ...
Find the balance in the account after 10 years
10 years is 40 quarters
a_40=2500(1+0.08/4)^40=$5520.10
a_1=34000
d=2250
a_n=34000+2250(n-1)
(a)a_5=34000+2250(4)=$43000
(b)S_5=(5/2)(34000+43000)=$192,500
A company buys a fleet of six vans for $120,000. During the next 5 years, the fleet will depreciate at a rate of 30% per year.
(a) Find the formula for the nth term of a geometric sequence that gives the value of the fleet t full years after it was purchased.
(b) Find the depreciated value of the fleet at the end of 5 full years.
a_1=120,000
r=1-0.3=0.70
(a) a_t=120000(0.7)^(t-1)
(b) a_5=120000(0.7)^4=$28,812
Find the coefficient with of the term
x^3y^5
in the expansion of
(3x+2y)^8
x^3y^5=x^(n-r)y^r
r=5
n=8
._8C_5=56
56*(3x)^3(2y)^5=56(27x^3)(32y^5)=48384x^3y^5
48384
Find the number of distinguishable permutations of the group of letters.
C,A,L,C,U,L,U,S
Use permutations but divide by the repeat letters
n=8
r=8
2 C's, 2 U's, 2 L's
(._8P_8)/(2!2!2!)=(8!)/8=(7!)=5040
A six sided dices is rolled six times. What is the probability that each side appears exactly once?
Total number of occurrences
6 possible numbers each time
n(S)=6*6*6*6*6*6=6^6=46656
Order does matter, so use nPr
Number of Events
n=6
r=6
n(E)=_6P_6=(6!)=720
(n(E))/(n(S))=720/46656=5/324
Another way to look at it is
The first roll, you have a 6/6 chance at selecting a number for the first time
The second roll you have a 5/6 chance at selecting a number for the first time
The third roll you have a 4/6 chance at selecting a number for the first time.
So on and so forth
6/6*5/6*4/6*3/6*2/6*1/6=5/324