Sums of arithmetic and geometric series

13-3 sums of arithmetic and geometric series

A series is a list of numbers that are added together.

ex: 2 + 4 + 6 + 8..

-To find the sum of the __ terms of an arithmetic series, use this formula:

sn= n(t1 + tn)/2

-To find the sum of the __ terms of a geometric series, use this formula:

sn= t1(1-r^n)/1-r

Ex problem 1:

Find the sum of the first 10 terms of the geometric series.

2 - 6 + 18 - 54 +

S10 = 2 (1- (-3)^10)/1-(-3) = 2(1-59049)/4

S10= -29524

Ex problem 2: All multiples of 4 between 2 and 75

4, 8 … + 72

tn= t1 + n-1 (d)

72= 4 + (n-1)4

72= 4+4n-4

The 4 and (-4) cancel out, leaving you with 72=4n.

Solve for n by dividing 72 by 4 and your answer is n=18

Logarithms

1st- If M and N are positive real numbers and b is a positive number other than 1,log b MN= log b M+ log b N.

2nd- is if log b M/N=log b M-log b N.

3rd- is log b M= log b N if and only if M=N.

4th- log b M^k=k log b M for any real number k. Use this rule when your trying to solve an exponent.

Example 1:

Expand:

ln x - (1/2) ln (x² + 1)

Example 2:

Condense: 3 log x + 2 log y - (1/2) log z

Example 3:

Expand:

Log₅ xy² =

Log₅ x + Log₅ y² = Log₅ x + 2 Log₅ y

Example 4:

Write as single:

Log₂ (x + 1) + Log₂ 3 = 4

Log₂ 3(x + 1) = 4

3(x + 1) = 2⁴

3x + 3 = 16
3x = 13
x = 13/3

Example 5:

Expand:

Log₇ (xy/z²) =
Log₇ x + Log₇ y - 2 Log₇ z

Example 6:

Write as single:

1) Log x + Log y - Log z =

Log (xy)/z

2) 2 Ln x + 3 Ln y =

Ln x²y³

3) (1/2) Ln x - (1/3) Ln y =

Logs

You can write a log as an exponent, and an exponent as a log.

logb(x) = a then, b^a=x

(b will be a subscript)

To solve a log you write as an exponent.

If no base is written for log, it is implied to be 10.

The base of ln is always e, and is worked the exact same as log.

If b and x are equal, than they cancel along with the log.

Ex 1:

Write the equation in exponential form.

log6(1/36)=-2

6^-2=1/36

Ex 2:

Write the equation in exponential form.

log4(16)=2

4^2=16

Ex 3:

Find each logarithm without using a calculator.

log3(27)

3^x=27

x=3

Ex 4.

Find each logarithm withougt using a calculator.

log4(64)

4^x=64

x=3

Ex 5:

Find the logarithm without using a calculator.

log10,000

10^x=10,000

x=4

Logs!

Helllo,
I hope everyone had a fantastic weekend! :)

1. Logs are another way to write exponents: logbx=a, as an exponent b^a=x
2. To solve a log you write as an exponent.
3. If no base is written for log, it is implied to be 10.
4. The base of ln is always e.
5. If b and x are equal, than they cancel along with the log.

Example 1:
Write the equation in exponential form.
log416=2
4^2=16

Example 2:
Write the equation in exponential form.
log6(1/36)=-2
6^-2=1/36

Example 3:
Write the equation in exponential form.
log101000=3
10^3=1000

Example 4:
Find the logarithm without using a calculator.
a) log100
=log10100=x
10^x=100
x=2
b) log10,000
=log1010,000=x
10^x=10,000
x=4
c) log0.01
log100.01=x
10^x=1/100
x=-2

Example 5:
Find each logarithm without using a calculator.
a) log50.2
3^x=9
3^2=9
x=2
b) log327
3^x=27
x=3
c) log3243
3^x=23
x=5
d) log33^8
3^x=3^8
x=8

Example 6:
Find each logarithm withougt using a calculator.
a) log464
4^x=64
x=3
b) log4(1/64)
4^x=1/64
x=-3
c)log4 fourth root of 4
4^x=fourth root of 4
4^x=4^1/4
x=1/4
d) log 41
4^x=1
x=0

Logs!

1. logₐMN = logₐM + logₐN

2. logₐM/N = logₐM - logₐN

3. logₐM = logₐN if and only if M = N

4. logₐM^k = klogₐM for any real number k. use when solving an exponential.

_________________________________________________________________________

Ex.1: Expand:

log XY^2=

log X+ 2log Y

Ex.2: Condense:

log 6- log 5+ log 2=

log 12/5

Log laws

There are 4 different laws for logarithms.  If M and N are positive real numbers and b is a positive number other than 1, then the first rule we have is log b MN= log b M+ log b N.  The second rule is if log b M/N=log b M-log b N.  The third rule is log b M= log b N if and only if M=N.  The last rule you need to know is log b M^k=k log b M for any real number k.  Use this rule when your trying to solve an exponent.

ex: log 5 2+ log 5 3= log 5 2*3= log 5 6
This is the first rule you need to know.  When you have addition of the logs, you multiply them together and get your answer.

ex:  log 18- log 2= log 18/log 2= log 9
This is the second rule you need to know.  When you have subtract of the logs, you divide the logs and solve for your answer.

Laws of Logarithms

If M and N are positive real numbers and a is a positive number other than 1 then:

1. logₐMN = logₐM + logₐN

2. logₐM/N = logₐM - logₐN

3. logₐM = logₐN if and only if M = N

4. logₐM^k = klogₐM for any real number k. use when solving an exponential.

Ex 1. log(MN)^2 = 2logM + 2logN

Ex 2. logM/N^2 = logM - 2logN

Ex 3. log M^3N/2 = 3logP + logN - log2

Ex 4. logMsqrtN/B = logMN^1/2 / B

= logM + 1/2logN - logB

Ex 5. log1/M = log1 - logM

Ex 6. log2 + log3 + log4 = log25

Ex 7. log8+ log5 - log4 = log40/4 = log10

Ex 8. 1/2log9 + log5 = log3 + log5 = log15

interest and stuff

A(t)=Ao(1+r)^t
-Used when compounding.
Ao=starting amount
r=rate
t= amount of time compounding in given time span.

A(t)=A0b^t/k
-Used when things are doubling, tripling, halving, etc.
Ao=starting amount
b=doubling, tripling, etc.
t=amount of time
k=how long it takes to double, triple, half, etc.

Rule of 72- to find how long it takes for something to double then find 72/r
r is not a decimal here

**Only used when compounding continuously
P(t)=Poe^rt
Po=starting amount
e=on ln button
r=rate
t=time

Example 1:
If $1,000 is invested in a savings account that is compounding continuously at .8% for 6 years how much money did you gain?
Compounding continuously
Pₒ = $1,000
e = e
r = .008
t = 72 months
P(t) = 1000(e^(.008*72))
= $1,778.91 – 1,000 = $778.91

Example 2:
The half-life of a radioactive isotope is 4 days. If 3.2 kg are present now, how much will be present after:
a) 4 days
A(t)=Aob^t/k
A(t)=3.2(1/2)^(4/4)
=1.6kg

b) 8 days
A(t)=Aob^t/k
=3.2(1/2)^(8/4)
=0.8 kg

c) 20 days
A(t)= Aob^t/k
A(t)=3.2(1/2)^(20/4)
=0.1 kg

d) t days
A(t)=Aob^t/k
A(t)=3.2(1/2)^t/4

Example 3:
How long does it take for any given amount to double at a rate of 8% increase per year?
This is a rule of 72 problem.
72 ÷ 8 = 9 years

Review degrees, radians, coterminal angels

Degrees is the form of numbers sometimes used in trigomonmetry. Degrees can be broken down inot minutes (') and seconds (")
1.To convert to minutes you take what is behind the decimal and multiply by 60.
2.To convert to seconds you take what is behind the decimal in new number and multiply by 60. (This only happens if there is another decimal. Sometimes there is not)

*To convert minutes and seconds back to degrees use formula
# + minutes/60 + seconds/3600

Radians are the perffered way to measure angles. All formulas require radians.
1.To convert degrees to radians use formula #*pi/180

*To convert radians back to degrees use formula Radians*180/pi

Coterminal angles are like non-reduced fractions. It is a more simple or more complicated name for an angle.
1.Add or subtract 36o n = coterminal angle (n can be any whole #)
2. For radians add or subtract 2pi n = coterminal angle (n can be any whole #)

Ex1
Convert 16.33 to degrees minutes and seconds
.33*60=19.8
.8*60=48
16degrees 19' 48"

Ex2
Convert 60 degrees to radians
60/180 = 1/3 = pi/3
Put pi because when using formula it belongs there.

Logs and Exponents

-You can use logs to write exponents. 

Written as log: logbx=a

Written as exponent= b^a=x

-When solving a log, you write it as an exponent

Example 1: log3 5= 56

3^56=5

Example 2: log2 3=6

2^6=3

Example 3: log7 9=12

7^12=9

-If the base is written for log, its 10. 

-The base of ln is always e

-The log cancels, if the base and other number are the same. 

Example 4: log3 3^2

log3 3^2=2

Logs

I did my blog today because i have a volleyball tournment sunday and i won't feel like doin it then. ha ha.

1. Logs are another way to write exponents: logb x = a
As an exponent b^a = x
2. To solve a log you write as an exponent
3. If the base is written for log it's implied to be 10
4. The base of ln is always e
5. If b and x are equal then they cancel along with the log

Example 1: log4 64 = 3
4^3 = 64

Example 2: log5 1/125 = x
5^x = 1/125
x = -3

Example 3: log2 2^8
the log cancels, the 2 cancels, and the other 2 cancels
your answer being = 8

--Daniellee

Exponents!:)

1. b^x * b^y = b^ x + y

2. b^x / b^y = b^ x-y

3. any number raised to the 0 = 1

4. (ab)^x = a^x b^x

5. (a/b)^x = a^x/b^x

6. (b^x) ^y = b^xy

7. nth root of a = a^1/n

8. nth root of a^y = a^y/n

9. x^-n = 1/x^n

10. a * b^x = cant simplify

11. (-a)^2 = positive number

12. -a^2 = negative number

    
    

    
    

    Example 1.) 6^0=1

    Example 2.) x^2 * x^3=x^5

Logs

Logs.

Logs look like this: logb x = a

As an exponent, it's b^a = x

b is the base of the exponent.
If there isn't any specific base, it's automatically 10.

If b and a are equal they cancel out and so does the log.

ln is a natural log; its base is automatically 10.

The exponent of a log can be written in front if you really feel like it.

Example:

log 4^5 = 6 could be written as 5log 4 = 6.

logs

log2 4=x
2^x=4
x=2

To solve a log, you write it as an exponent.  The standard log is logb x=a. To write it as an exponent, it is b^a=x.  Your b is your base, your a is your exponent, and the x is the number that the base and exponent are set equal to each other.  If no base is written for the log, it is implied to be 10.  If we have an ln, the base will always be e.  If b and x are equal then they cancel along with the log.  So for this problem, we have the base of the log as 2 and the exponent is x.  It is equal to 4, so we find common bases, which is 2.  Now we have 2 raised to the x and 2 raised to the 2.  Now you set the exponents equal to each other and solve for x and the answer ends up being x equals 2. 

Exponents!

1. b^x * b^y = b^ x + y

2. b^x / b^y = b^ x-y

3. any number raised to the 0 = 1

4. (ab)^x = a^x b^x

5. (a/b)^x = a^x/b^x

6. (b^x) ^y = b^xy

7. nth root of a = a^1/n

8. nth root of a^y = a^y/n

9. x^-n = 1/x^n

10. a * b^x = cant simplify

11. (-a)^2 = positive number

12. -a^2 = negative number

    _______________________________________________

    Ex.1: (-4)^-2

    1/16

    Ex.2: ((x^-2)^-1)^0

    The answer will be 0 because everything is raised to 0

Exponents

When multiplying numbers you add exponents.

When dividing exponents you subtract.

When any number is raised the zero power it equals one.

When (ab)^x and (a/b)^x distribute the exponent to each number.

When (b^x)^y multiply exponents.

When you have a cubed root etc. of a number it becomes a^1/n.

When a number is raised to a negative exponent bring it to the bottom.

Example 1.

Simplify.

1. (-4)^-2

Since the exponent is negative drop it down and you now have a fraction. Since the number is parentheses its going to be a positive number and 4 squared is 16. So the answer is 1/16.

2. ((x^-2)^-1)^0

Since the whole problem is raised to the zero exponent the answer is 0.
3. -4^-2

Since the exponent is negative drop it down and you now have a fraction. Since the number is not in parentheses its going to be a negative number and 4 squared is 16. So the answer is -1/16.

Formulas!

Hello everyonee,
I hope everyone had a wonderful weekend!

A(t)=Ao(1+r)^t
-Used when compounding.
Ao=starting amount
r=rate
t= number of times compounding in given timespan.

A(t)=Aob^t/k

-Used when things are doubling, tripling, halfing, etc.

Ao=starting amount

b=doubling, tripling, etc.

t=amount of time

k=how long it takes to double, triple, half, etc.r.

Rule of 72- to find how long it takes for something to double then find 72/

r is not a decimal here

**Only used when compounding continuously

P(t)=PoE^rt

Po=starting amount

e=on ln button

r=rate

t=time

Example 1:

The half-life of a radioactive isotope is 4 days. If 3.2 kg are present now, how much will be present after:

a) 4 days

A(t)=Aob^t/k

A(t)=3.2(1/2)^(4/4)

=1.6kg

b) 8 days

A(t)=Aob^t/k

=3.2(1/2)^(8/4)

=0.8 kg

c) 20 days

A(t)= Aob^t/k

A(t)=3.2(1/2)^(20/4)

=0.1 kg

d) t days

A(t)=Aob^t/k

A(t)=3.2(1/2)^t/4

Example 2:

a)If $1000 is invested so that it grows at the rate of 10% per year, what will the investment be worth in 20 years?

A(t)=Ao(1+r)^t

A(t)=1000(1+.10)^20

=$6727

b)According to the rule of 72, in approximately how many years will the investment double in value?

72/10=7

Compounding

You use the following formula when compounding: A(t)=Ao(1+r)^t

Ao=starting amount

R=rate

T=amount of time compound in given time

You use the following formula when doubling, tripling, halfing, etc.: A(t)=Aob^(t/k)

Ao= starting amount

B=doubling, tripling, halfing, etc. 

T=amount of time

K=how long it takes to double triple, etc. 

To find out how long it takes for something to double use the Rule of 72: 72/r

R=rate 

When compounding continuously, you use: P(t)=PoE^(rt)

Po=starting amount

E=on ln button

R=rate

T=time

Example 1: How long does it take for a given about to double at a rate of 3% increase per year? 

72/3=24 years

Example 2: How long does it take for an amout to double at a rate of 60% increase per year? 

72/60=1.2 years

Chapter 5-3 and 5-4

Hope everyone that went to prom/prom mania had a great time!!

Formulas:

1. A(t) = A0(1+r)^t - used when compounding
A0 - starting amount
r - rate
t - number of times compounded in given times span

2. A(t) = A0b^t/k - used when things are doubling, tripiling, halfing, ect
A0 - starting amount
b - doubling, tripiling, , ect
t - amount of time
k - how long it takes to double, triple, half, ect

3. Rule of 72 - to find how long it takes for something to double. Find 72/r
*r is not a decimal here

4. **Only use when compounding continuously
P(t) = P0e^rt
P0 - starting amount
e - on ln button
r - rate
t - time

Example 1: A bank advertizes that if you open a savings acconunt you can double your money in 12 years. If you invest $1,000, how much money will you have after 5 years?

You use formula 2 because its talking about double and thats the only formula that talks about doubling.
A = A0b^t/k
A(t) = 1,000(2)^5/12
= $1,334.84

Example 2: If $1,000 is invested in a savings account that is compounded monthly at .5% for 5 years how much money did you gain?
b. What if it was compounded continuously?

A(t) = A0(1+r)^t
t = 6 months
A(t) = 1,000(1+.005)^60 = $1,348.85
b. P(t) = P0e^rt
= 1,000e^(.005)60
= $1, 349.86

--Daniellee

Compounding Interest, Compounding Continuously, Rate, and etc.

1. A(t) = Aₒ(1 + r)^t

• Used when compounding

• Aₒ = starting amount

• r = rate

• t = amount of time compound in given timespan

2. A(t) = Aₒb^(t/k)

• Used when things are doubling, tripling, halfing, etc.

• Aₒ = starting amount

• b = doubling, tripling, etc.

• t = amount of time

• k = how long it takes to double, triple, etc.

3. Rule of 72 → 72 ÷ r

• Used to find how long it takes for something to double

• r is the rate. Not used as a decimal, but a percent.
  
  
  

4. P(t) = Pₒe^(rt)

   only used when compounding continuously

• Pₒ = starting amount

• e = on ln button

• r = rate

• t = time
  
  
  

Ex 1. A bank advertises that if you open a savings account, you can double your money in 10 years. If you invest $2,000, how much money will you have after 3 years?

It gives you how long it takes to double, so you use formula 2.

• Aₒ = $2,000

• b = 2

• t = 3

• k = 10

A(t) = 2,000(2)^3/10

= $2462.29

Ex 2. If $1,000 is invested in a savings account that is compounding monthly at .8% for 6 years how much money did you gain? What if it was compounding continuously?'

First, we will use the compounding formula, then the compounding continuously formula for the second question. It is compounding monthly so we have to convert the years into months.

• Aₒ = $1,000

• r = .008

• t = 6*12 = 72

A(t) = 1,000(1 + .008)^72

= $1,774.84

They asked for how much gain so we will subtract the original amount from the answer we got. Amount gained was $774.84

Second part:

Compounding continuously

• Pₒ = $1,000

• e = ...e

• r = .008

• t = 72 months

  P(t) = 1000(e^(.008*72))

  = $1,778.91 – 1,000 = $778.91

Ex 3. How long does it take for any given amount to double at a rate of 8% increase per year?

This is a rule of 72 problem.

72 ÷ 8 = 9 years

Formulas

1. A(t) = Aₒ(1 + r)^t ---> Used when compounding

Aₒ = starting amount r = rate

t = amount of time compound in given timespan

2. A(t) = Aₒb^t/k ---> Used when things are doubling, tripling, halfing, etc.

Aₒ = starting amount b = doubling(2), tripling(3), etc.

t = amount of time k = how long it takes to double, triple, etc.

3. Rule of 72 – to find how long it takes for something to double

72 ÷ r% r is not a decimal

4. P(t) = Pₒe^rt only used when compounding continuously

Pₒ = starting amount e = on ln button

r = rate t = time

Ex 1. A bank advertises that if you open a savings account, you can double your money in 12 years. If you invest $1,000, how much money will you have after 5 years?

A(t) = 1,000(2)^5/12 = $1,334.84

Ex 2. A radioactive substance has a half-life of 5 days. This means that half the substance decays in 5 days. At what rate does the substance decay each day?

A(t) = Aₒ(1/2)^t/5 Aₒ = (1/2)^1/5 · t Aₒ = (.87)^t

1 - r = .87 r = .13 = 13%

Ex 3. If $1,000 is invested in a savings account that is compounding monthly at .5% for 5 years how much money did you gain? What if it was compounding continuously?

A(t) = 1,000(1 + .005)^60 = $1,348.85 $348.85

P(t) = 1,000e^.005(60) = 1,000e^.3 = $1,349.85 $349.85

Formulas

This fomulas are used to find how much a certain item will amount to after so much increase over a given amount of time. This is commonly used to for money and bunnies.

1.A(t)=A0 (1+r)^t.......................Used when compounding
A0=starting amount
r=rate
t=number of times compounding in given timespan

2.A(t)=A0 b ^t/k.........................Used things are doubling, halfing, tripling, etc.
A0=starting amount
b=double (2) or half (1/2) or triple (3), etc.
t=amount of time
k=how long it takes to double, half, triple , etc.

3.Rule of 72. To find how long it takes for something to dougle then find 72-r (r is not a decimal in this situation)

4.P(t)-P0 e ^rt..............................only used when compounding continuously
P0=starting amount
r=rate
t=time
e=number e

1.The local bank offers acounts were you money is doubles after 10 years. If you deposit $1,200 how much money will you have in 8 years?
Use formula #2 because you are doubling.
Ao=1200
b=2 (because your doubling)
t=8 (the amount of time you plan to leave money in account)
k=10 (time it takes to double)
A(t)=1200(2)^8/10=2089.32

Logs

Logs are another way to write exponents
logbx=a
as an exponent b^a=x
To solve a log you have to write it as and exponent
If the log has no base than its base is 10
The base of ln is e
If b and x are equal than they cancel with the log
log M N = log M + log N
log M/N = log M – log N
log M = log N if M = N
log M^k = k log M
if there’s a log and you have to expand all the stuff in the numerator is added and all the stuff in the
denominator is subtracted
Examples
Rewrite in exponential form
1) log 1000 = 3
10^3 = 1000
2) ln 8 = 2.1
e^2.1 = 8
Expand
3) log (MN)^2
2log M + 2 log N
4) log M/(N^2)
log M – 2log N
5) log (M^2)N/(P^3)
2log M + log N – 3log P

Exponets

1. b^x * b^y = b^ x + y

2. b^x / b^y = b^ x-y

3. any number raised to the 0 = 1

4. (ab)^x = a^x b^x

5. (a/b)^x = a^x/b^x

6. (b^x) ^y = b^xy

7. nth root of a = a^1/n

8. nth root of a^y = a^y/n

9. x^-n = 1/x^n

10. a * b^x = cant simplify (pemdas)

11. (-a)^2 = positive number

12. -a^2 = negative number

Example 1

Example 2

exponents

Exponents

When multiplying numbers you add exponents.

When dividing exponents you subtract.

When any number is raised the zero power it equals one.

When (ab)^x and (a/b)^x distribute the exponent to each number.

When (b^x)^y multiply exponents.

When you have a cubed root etc. of a number it becomes a^1/n.

When a number is raised to a negative exponent bring it to the bottom.

Example 1.

Simplify.

1. (-4)^-2

Since the exponent is negative drop it down and you now have a fraction. Since the number is parentheses its going to be a positive number and 4 squared is 16. So the answer is 1/16.

2. ((x^-2)^-1)^0

Since the whole problem is raised to the zero exponent the answer is 0.
3. -4^-2

Since the exponent is negative drop it down and you now have a fraction. Since the number is not in parentheses its going to be a negative number and 4 squared is 16. So the answer is -1/16.

Sums Of Arithmetic & Geometric Series

To find the sum of the _____ terms of an arithmetic series.

Formula: Sn= n(T1+Tn)/2

To find the sum of the _____ terms of a geometric series.

Formula: Sn= T1(1-r^n)/1-r

Series is a list of numbers that are added together.

Ex1: 2+4+6+8....

Ex2: Find the first 25 terms of the arithmetic series

11+ 14 + 17 + 20....

(This formula learned in chapter 13-1)

Tn = T1+ (n-1)d

Tn = 11+ (25-1) 3

Tn= 11+ 75 -1

Tn=83

S25= 25(11+83)/2

=1,175

1)Plug into formula learned in 13-1

2)Plug the Tn you get from that formula into the new formula.

3) Once plugged into formula and calculator, you get 1,175

Exponents

This past week we've recently learned about exponents. There are many different steps you must follow in order to correctly solve these problems.

Steps:

1. b^x * b^y = b^ x + y

2. b^x / b^y = b^ x-y

3. any number raised to the 0 = 1

4. (ab)^x = a^x b^x

5. (a/b)^x = a^x/b^x

6. (b^x) ^y = b^xy

7. nth root of a = a^1/n

8. nth root of a^y = a^y/n

9. x^-n = 1/x^n

10. a * b^x = cant simplify (pemdas)

11. (-a)^2 = positive number

12. -a^2 = negative number

EX 1

(-4)^-2 = ?

• Since the 4 is in parentheses, the number will be positive.

• Because the number is raised to a negative exponent, you would take the reciprocal of 4^2, which is 1/16

• Your answer is 1/16.

EX 2

(3n^2) ^-1 (3n^2) ^7 = ?

• 3 ^ -1+ 3 ^ 7= 3^6

• n^-2+14= n^12

• Your answer is 3^6n^12

Limits

-If it does not follow the rules

1. Type the equation into y= in calculator

2. Hit 2nd table

3. Type in numbers 10, 100, 1000, 10000. Do more numbers if needed to see what number the equation is approaching.

**Make sure your calculator is in ask**

-If it is geometric (has #^n)

1. If r <1 = 0

2. If r >1= + or - infinity

Examples: Find the limit of

A: (x^3+x+10)/(x^5+8)

The bottom degree is bigger so it will eventually go to zero to 0.

The limit = 0.

B: (2)^n

It’s geometric and r > 1 so the limit is infinity.

C. (x+20)/(x-20)

Follows polynomial rules, degrees are equal. Coefficents are 1 and 1, so the limit is 1/1 = 1.

D. (x^9)/(x)

Top exponent is larger so the limit is infinity. It is positive infinity because when you plug a number in, it comes out positive.

Limitsssss

Infinite Limits can go on forever the bigger the number is.

-If you have a fraction with 2 polynomials divided use the these rules: (Refers to exponents)

1. Degree of top = degree of bottom leading coefficient/leading coefficient

2. Degree of top > degree of bottom + or - infinity (plug into equation to see if + or -)

3. Degree of top < degree of bottom 0



If it isnt following the rules then you...

1. Type the equation into y= in calculator

2. Hit 2nd table

3. Type in numbers 10, 100, 1000, 10000. Do more numbers if needed to see what number the equation is approaching.



*If it is geometric (has #^n)

1. If r <1 = 0

2. If r >1= + or - infinity



Examples: Find the limit of

A: (b+1)/b

The exponents equal eachother so the it is 1
The limit is 1

B:
Lim n -> k^2+3/2k^2-3
Top and bottom are equal so the limit is 1/2

Arithmetic and Geometric Series!

To find the sum of the ___ terms of an arithmetic series

Sn=n(t1+tn)/2

To find the sum of the ___ terms of a geometric series

Sn=t1(1-r^n)/1-r

Series- a list of numbers that are added together.

1+3+5+9+...

______________________________________________________________________

Ex.1:

Find the sum of the series 1-3+5-7+9-11+...+1001.

S501=501(1+1001)/2

S501=251001

1001=1+(n-1)2

1001=1+2n-2

1001=-1+2n

1002=2n

n=501

Ex.2:

For the arithmetic series, find the specified sum.

S50=: 5+10+15+...

S50=50(5+250)/2

S50=6375

tn=t1(n-1)d

tn=5+(50-1)5

tn=5+250-5

tn=250

Exponents

This is a blog about exponents.

Exponents look like this: x^n.

To simplify an exponent, you multiply x n times.

Here's some rules:

When you multiply, you add exponents.

x^n * x^m = x^(n+m)

When you divide, you subtract exponents.
x^n/x^m = x^(n-m)

Distributin them exponents:

(x*y)^n = x^n * y^n

Also in division:

(x/y)^n = x^n/y^n

And when you raise it to the 0 power, you get 1. Always.

(pi+687698757865897698578658768976-2+6565646*767676767/9-2)^0 = 1

There's some other stuff too but I'm pretty sure everybody else took notes too.

Arithmetic and Geometric Series

Hellooooo people, hope all of you had a terrific weekend!

To find the sum of the ___ terms of an arithmetic series

Sn=n(t1+tn)/2

To find the sum of the ___ terms of a geometric series

Sn=t1(1-r^n)/1-r

Series- a list of numbers that are added together.

2+4+6+8+...

Example 1:

For the arithmetic series, find the specified sum.

S50=: 5+10+15+...

S50=50(5+250)/2

S50=6375

tn=t1(n-1)d

tn=5+(50-1)5

tn=5+250-5

tn=250

Example 2:

Find the sum of all multiples of 3 between 1 and 1000.

3+6+...999

S333=333(3+999)/2

S=166833

999=3+(n-1)-3

999=3n

n=333

Example 3:

Find the sum of all positive 3-digit numbers whose last digit is 3.

103+113+...993

S90=90(103+993)/2

S90=49320

tn=t1+(n-1)d

993=103+(n-1)10

993=103+10n-10

993=93+10n

900=10n

n=90

Example 4:

Find the sum of the series 1-3+5-7+9-11+...+1001.

S501=501(1+1001)/2

S501=251001

1001=1+(n-1)2

1001=1+2n-2

1001=-1+2n

1002=2n

n=501

----Danielleeee

Arithmetic and Geometric Series and Their Sums!

Hello everyone,

I hope everyone had a wonderful weekend!

To find the sum of the ___ terms of an arithmetic series

Sn=n(t1+tn)/2

To find the sum of the ___ terms of a geometric series

Sn=t1(1-r^n)/1-r

Series- a list of numbers that are added together.

2+4+6+8+...

Example 1:

For the arithmetic series, find the specified sum.

S10: t1=3, t10=39

S10= 10(3+39)/2

S10=210

Example 2:

For the arithmetic series, find the specified sum.

S50=: 5+10+15+...

S50=50(5+250)/2

S50=6375

tn=t1(n-1)d

tn=5+(50-1)5

tn=5+250-5

tn=250

Example 3:

Find the sum of all multiples of 3 between 1 and 1000.

3+6+...999

S333=333(3+999)/2

S=166833

999=3+(n-1)-3

999=3n

n=333

Example 4:

Find the sum of all positive 3-digit numbers whose last digit is 3.

103+113+...993

S90=90(103+993)/2

S90=49320

tn=t1+(n-1)d

993=103+(n-1)10

993=103+10n-10

993=93+10n

900=10n

n=90

Example 5:

Find the sum of all positive 3-digit numbers divisible by 6.

102+108+...996

S150=150(102+996)/2

S150=82350

996=102+(n-1)6

996=102+6n-6

996=96+6n

900=6n

n=150

Example 6:

Find the sum of the series 1-3+5-7+9-11+...+1001.

S501=501(1+1001)/2

S501=251001

1001=1+(n-1)2

1001=1+2n-2

1001=-1+2n

1002=2n

n=501