Reflection

This week we finished learning about chapter 13. In chapter 13 we learned about sequences, many formulas, sigma notation, etc.
Sigma Notation

100 - 1 & 100 are called the limits of summation.
sigma n^2 - n^2 is the summand
n=1 - the bottom variable is called the index, which is K.

- To evaluate-plug in the numbers between your limits of summation into the summand.
Adding each term to form a series.

- Examples:
Give each series in expanded form

1) 4
sigma 5K = 5+10+15+20
K=1

2) 6
sigma 9+16+25+36
n=3

Express the series in sigma notation

1) 1+2+4+8+16+32
5
sigma 1X(2)^K
K=0

2) 48+24+12+6+...
infiniti
sigma 48(1/2)^c
c=0

For some reason i am still having trouble with sequences even though it super easy, i guess i need to know the formulas better.

REFLECTION #25

Well this week went by reaaaaallly fast! which was awesome ha. Starting off this week we learned infinite sequences and series with limits. And then we learned how to find the sums of infinite series. Also we learned how to write repeating decimals as fractions which was super easy. Then on Wednesday we learned sigma notation. Everything we learned this week was pretty easy so i'll give an example of each thing we learned.

Ex. 1.) Find the given limit. > lim 3n^2 + 5n/8n^2
*First looking at this problem you see that it's a fraction with polynomials so you can use the rules in your notes to find the limit of this.
*You would use the first rule because the degree of the top equals the degree of the bottom.
*So the limit would be the coefficients.
*So you get 3/8 as your answer

Ex. 2.) Find the given limit. > lim tan(1/n)
*This problem does not relate to any of the rules so we have to plug in numbers to find out what the limit is. (First you plug in 100, then 1000, then 10000)
*So you get tan(1/100) = approximately .010
*Then you get tan(1/1000) = approximately .001
*Then you get tan(1/10000) = approximately .0001
*Then you ask yourself, "What number is it approaching?"
*Based on the answers you got, it looks that the numbers are getting closer and closer to zero so your answer is zero.

Ex. 3.) Find the sum of the infinite series: 1+1/2+1/4+1/8+...
*The first thing you have to do is find out whether the series is arithmetic or geometric
*It is geometric because you can divide 1/8 by 1/4 and get 1/2, then divide 1/4 by 1/2 and get 1/2. So you know your ratio is 1/2.
*Then all you have to do is use the formula>> S = t1/1-r
*So you get S = 1 / (1-1/2)
*And your answer is 2.

Ex. 4.) Convert each repeating decimal to a fraction:

a.) 0.77777...
*For this one, only the 7 is repeating and it's in the tenths place.
*So you use the formula 7/(10-1). (*since "10" means the tenths place.)
*And you get 7/9

b.) 0.636363...
*In this one, the 63 is repeating and it's in the hundreths place.
*So you plug into the formula and get > 63/(100-1) and that gives you 63/99
*And that reduces to 7/11

Ex. 5.) Write in expanded form: (yeah let's pretend the E is a sigma..ha i think you get the point)
7
E (4n-7)
n=3

*Okay in this problem you are looking for the numbers in a series when you plug in the numbers 3 through 7 into (4n-7). (if that makes any sense..)
*So you plug in 3 first and you get 4(3)-7 which equals 5. So 5 is your first number in the series.
*Then you plug in 4 and you get 9 as the next number in the series.
*Then you plug in 5 and you get 13 as the next number.
*Then you plug in 6 and get 17, and plug in 7 and get 21.
*So your expanded series looks like this: 5+9+13+17+21

Ex. 6.) Write in sigma notation: 4+8+12+16+20
*First you have to find out if the series is arithmetic or geometric.
*It's arithmetic because you add 4 each time.
*So you plug it into the formula tn=t1+(n-1)d
*And you get tn=4+(n-1)(4)
*So tn = 4n (*4n is the summand)
*to get the bottom variable, the index, you have to plug in a number into "4n" that will give you the first number in the series.
*So the index is n=1
*And to get the top number you count the numbers in the series starting with 1 in this case.
So the top number is 5.

5
E 4n
n=1

**Now for what I don't understand. I'm having trouble with numbers 33, 34, and 35 on page 509. I don't know, they're just confusing to me. So if anyone would like to explain that would be greatly appreciated (:

reflection

So this week is the easiest thing in the whole world because its the easiest thing in the united states of america. Im gonna give you the formulas and then im going to show you how to use them.

Arithmetic formula:

tn = t1 + (n-1)d
n = term #
t1 = first term
d = what you add


Geometric to find a term

tn = t1 x r^(n-1)
r = what you multiply by

NOW TO USE THEM:

Arithmetic Sequence:
5,10,15... find the 12th term.

5+11(5)

60

Geometric Sequence:
2,4,8,16.... find the 6th term

2 x 2^5

32!

now you know how to use it:)

So this week we started learning Chapter 13. If you memorize the formulas, it helps a lot. Here’s some stuff from 13-2 and 13-3
Example:Find the recursive definition of 81, 27, 9, 3
an=an-1/3

Find the recursive definition of 1, 2, 6, 24, 120, 720
n=1 1
n=2 2 an-1Xn
n=3 6
n=4 24

Arithmetic Sn=n(t1+tn)/2
Geometric Sn=t1(1-r^n)/1-r

Example:Find the sum of the first 10 terms of the series.
2-6+18-54+... geometric
r=-6/2=-3
S10=2(1-(-3)^10)/1-(-3)= -29,524

something that I didn’t understand was in chapter 13 section the last example in our notes I don’t understand where the 3 comes from.

Reflection 24?

I can't remember the number but i'm guessing it's 24!
wel, carnival ball was great last night & this week went by extremely fast, so that was good. :)

we worked on Sequences and Series.

let's start with formula review:

arithmetic sequence:

tn=t1+(n-1)d

*d being your ratio, or number that's added or subtracted.
*t1 being your first term
*n being the term in the sequence we're trying to find


ex:
In the arithmetic sequence:3,5,7,9-- find the 28th term.

t28=3+(27)(2)
t28=3+54
t28=57

geometric sequence, or a sequence with multiplication and division:

tn=t1*r^(n-1)

*t1 is first term.
*r is your ratio, multiplied or divided by.
*missing term.

ex:
In the geometric sequence: 2,4,8,16-- find the 10th term

t10= 2*2^9
t10= 2*512
t10= 1024

---------------------
i'm still struggling with the last section we learned...

4/11

this week was pretty fun haha since it was spring break and all... :/ haha time for school.

sequences and series:

arithmetic series: 3,5,7,9...
geometric series: 2,4,8,16...

arithmetic formula: a.n = a.1 + (n-1)d
geometric formula: a.n = a.1 x r^(n-1)

you can solve for any term number by plugging in the numbers to the formula, its not that difficult and i liked this chapter

reflection 24?

Ok, so this week was fairly easy! except for that test that i bombed! But arithmetic and Geometric sequences are easy! woop woop

Arithmetic formula:

tn = t1 + (n-1)d
n = term #
t1 = first term
d = what you add


Geometric to find a term

tn = t1 x r^(n-1)
r = what you multiply by

EXAMPLES:

Arithmetic Sequence:

1, 4, 2, 5, 3.....

Geometric Sequence:

1, 2, 6, 24, 120...

___________________________________________________

Now, i dont understand how to find recursive definitions. I dont have a clue on anything about them!

1/31/10

We learned sequences this week

2main types:
arthimetic-where you add or subtract
geometric-where you multiply or divide

Formulas:
Arithmetic to find a term
tn=t1+(n-1)d n=term number t1=first term
d=what you add
tn=term number________actually in a sequence

Geometric tn=t1xr(n-1)
r is what u multiply by

Recrusive definition-formula for a sequence that involves the previous term (an-1) or tn-1

Find the recrusive definiton for 81,27,9,3
an=an-1/3


Arithmetic sums and geometric sums
Arithmetic=sn=n(t1+tn)/2
Geometric sn=t1(1-rn0/1-r

Reflection #24

okay, this week was not too bad, but i do not really understand as much as i did last week. haha, i only understand it when i do my homework, or when i am in class. but when it comes to the quiz, i blank and fail every quiz. i don't know why i do that all of the time, but i do it on every chapter.
__________________________________________________________________________


Arithmetic to find a term
a.n = a.1 + (n-1)d

n = term #
a.1 = first term
d = what you add, or difference
a.n = term #___ actually in sequence

Geometric to find a term
a.n = a.1 x r^(n-1)


r = what you multiply by

_________________________________________________________________________

Arithmetic example
Find the formula for the nth term of the arithmetic sequence.
3,5,7...
n = a.1 + (n-1)d
3000 = 3 + (3000-1)2

3000 = 6001

Geometric example
Find the formula for the nth term of the sequence.
3, 4.5, 6.75
r = 4.5/3 = 3/2

n = 3 x (3/2)^n-1
________________________________________________________________________

So, i am having some trouble in this chapter, and definitely need some help. If anyone wants to show me examples or explain how to do everyting, i would appreciate any help i can get. Because i do not fully understand this chapter. THANKSS:)

reflection 24

this week we started chapter 13: arithmetic and geometric sequences

Arithmetic

1,4,7,10
usually the problem asks for a certain term in the sequence...do not do this in your head...use the formula!
An= A1+(n-1)d

D=what you add
A1=first term

here is what you do for the sequence above

An= 1+(n-1)3
An=1+3n-3
-2+3n is the formula

Geometric
3,6,12,24,...

again, dont be lazy, use the formula

An = A1 x r^(n-1)
r=what you multiply by

An=3 x 2^(n-1)

Reflection 1/31

We learned about arithmetic and geometric formula this week in chapter 13, this is to find a certain number in a sequence of numbers.

Arithmetic formula:

a.n = a.1 + (n-1)d



Geometric formula:

a.n = a.1 x r^(n-1)

r = what you multiply by


An example of an arithmetic sequence is:

1,3,5,7....

An example of a geometric sequence is:

3.25, 5.5, 7.75, ...

To find these a certain number in these sequences you just use the formulas and plug in.

reflection 24

Chapter 13

FORMULAS:

*arithmetic - tn=t1 + (n-1)d

*geometric - tn=t1 x r^(n-1)


1) Find the formula for the nth term of the arithmetic sequence 3,5,7....

tn=3+(n-1)(2)
tn=3+2n-2
tn=1+2n


2) Find the formula for the nth term os the sequence 3,4.5,6.75....

r=4.5/3 = 3/2

6.75/4.5 = 3/2

tn=3 x (3/2)^n-1


3) In a geometric sequence t3=12 and t6=96. Find t11.

12r^3=96
r^3=8
r=2

r=2 t1=3

t11=3 x (2)^10 = 3072

Reflection numero 24

This week wasn't bad at all. I'ma crush me some chapta 13. Lets talk about some sequences:

Ok so if you got a sequence such as lets say: 2, 5, 8, 11,........ it is arithmetic because it increases by 3 each time. So the formula to find another term down the line would be:
t(n) = 2 + (n-1)3
because the formula for an arithmetic series is t(n) = t(1) + (n-1)d....t(1) means term one, n is the term number you are looking for, and d is the amount the series increases by.

-So if i wanted to find the 7th term of this sequence, if would plug in:
t(7) = 2 + (7-1)3
which equals 21. So the 7th term in this series is 21.


Now, the sequence 3, 4.5, 6.75,.... is a geometric sequence because it doesn't increase by the same number each time, it goes by a ratio in this case.
To find the ration, divide 4.5 by 3.....it equals 3/2. Then divide 6.75 by 4.5 jus to make sure...it equals 3/2. If it wouldn't have it would have been neither.

So now how do you find the 7th term in this sequence....by using this formula to start:
t(n) = t(1) x r^(n-1)
-r is your ratio.

So lets plug in an get our answer:

t(7) = 3 x 3/2^(7-1)

its gives you 2187/64, so thats your 7th term.

______________________________________________________________
i pretty much get everything, if someone wants to do an example of a reverse one they can...

Reflection #24

Arithmetic to find a term
a.n = a.1 + (n-1)d

n = term #
a.1 = first term
d = what you add, or difference
a.n = term #___ actually in sequence

Geometric to find a term
a.n = a.1 x r^(n-1)

r = what you multiply by


Arithmetic example
Find the formula for the nth term of the arithmetic sequence.
3,5,7...
a.n = a.1 + (n-1)d
a.3000 = 3 + (3000-1)2

a.3000 = 6001

Geometric example
Find the formula for the nth term of the sequence.
3, 4.5, 6.75
r = 4.5/3 = 3/2

a.n = 3 x (3/2)^n-1

Reflection #24

First, let's review all the formulas:

1) to find a term
arithmetic...tn = t1 + (n-1)d
where n=term #, t1=first term, d=what you add, tn=term you're looking for
geometric...tn = t1 x r^(n-1)
where r=what you multiply by
*is written...2, 3, 4, 5,...

2) to find the sum
arithmetic...sn = n(t1 + tn) / 2
geometric...sn = t1(1 - r^n) / 1-r
** tn-1 = previous term
***tn-2 = previous to the previous term
****is written...2 + 3 + 4 + 5 +...

Examples:
Find the formula for this arithmetic sequence: 3, 5, 7,...
tn = 3 + (n-1)2
tn = 3 + 2n - 2
tn = 1 + 2n

Find the next four numbers in the sequence: t1 = 1 & tn = 3(tn-1) - 1
1, 2, 5, 14, 41,...

Find the sum of the first ten terms of the series: 2 - 6 + 18 - 54 +...
s10 = 2(1 - (-3)^10) / 1 - (-3)
s10 = -29, 524

Find the sum of the first 25 terms of the arithmetic series: 11 + 14 + 17 + 20 +...
tn = 11 +(25 - 1)3
tn = 83
s25 = 25(11 +83) / 2
s25 = 1175

Now, what I don't understand is...#s 29 through 34 on page 489 (part of our homework)
I'm lost...I was also lost on #5 on the quiz we took Friday.
Any help?

Reflection 24

This week wasn't too bad, we started chapter 13. It's the same as usual just formulas to memorize. So far chapter 13 has been simple math. We learned the two types of sequences arithmetic and geometric. And also how to find formulas for sequences and how to find a certain number.

Recursive Definitions-formula for a sequence that involves the previous term (an-1) of tn-1

Example:Find the recursive definition of 1, 2, 6, 24, 120, 720

*The first thing you do is figure out if its arithmetic or geometric. If it's arithmetic it's either going to be addition or subtraction. And if it's geometric it's going to be divison or multiplication. If you can't figure out what the pattern is by just looking at it then you might have to do a chart like bellow.

*In this problem it's a little complicated the numbers are added diagonally to get the numbers on the right. (2x1=2,3x2=6,4x6=24)
n=1 1
n=2 2 an-1X
n=3 6
n=4 24

--------------------------------------------
Now for what I don't get, the section after this is a little confusing to me. And when you have to find the other terms, I don't really get that.

Reflection 24

This week went by extremely slow, but I'm glad it's over and we had a half day Friday. This week in Advanced Math we learned about sequences and series, different formulas dealing with those sequences and series, and the different types of them, such as arithmetic and geometric. Arithmetic sequences are when you have to add or subtract to get the next term(s). Geometric sequences are when you have to multiply to get the next term(s).

Here are some formulas and examples: period(.) denotes subscript

Arithmetic to find a term
a.n = a.1 + (n-1)d

n = term #
a.1 = first term
d = what you add, or difference
a.n = term #___ actually in sequence

Geometric to find a term
a.n = a.1 x r^(n-1)

r = what you multiply by


Arithmetic example
Find the formula for the nth term of the arithmetic sequence.
3,5,7...
a.n = a.1 + (n-1)d
a.3000 = 3 + (3000-1)2

a.3000 = 6001

Geometric example
Find the formula for the nth term of the sequence.
3, 4.5, 6.75
r = 4.5/3 = 3/2

a.n = 3 x (3/2)^n-1

REFLECTION #24

Well this week went by pretty fast if you ask me. Monday we review polar and rectangular for the test on Tuesday. Then on Wednesday we started Chapter 13 with sequences. I think it's pretty easy so far, but we'll see how it goes once we learn new things. It'll probably get harder.
Anyway, we learned the two types of sequences: arithmetic and geometric. And also how to find formulas for sequences and how to find a certain number in a sequence. Then on Thursday we learned recursive definitions and arithmetic sums and geometric sums.

Here's a few examples of what we learned this week:

Ex. 1.) Find a formula for tn: 17, 21, 25, 29,...
*First off, you have to determine what kind of sequence this is. Arithmetic or geometric? You can look at the numbers and see what is added between them, or if something could have been multiplied.
*In this case, 4 was added each time, so that makes this sequence an arithmetic sequence.
*So you will use this formula: tn = t1 + (n-1) d
*And you just plug in numbers from the sequence into the formula like this:
*t1 means the first term of the sequence. So t1 is 17. And "d" is the difference (meaning what is added each time). So d is 4. (And you just leave "n" as "n".)
*So you get tn = 17 + (n-1)(4)
*And you solve it like a regular equation by distributing the 4 and you get..
*tn = 17 + 4n - 4 Then subtract 4 from 17
*And your formula is tn = 13 + 4n

Ex. 2.) Find the 3rd, 4th, and 5th term in the sequence: t1=2; t2=4; tn=tn-1 + tn-2
*For this all you do is plug in what you know into the formula you are given.
*First you want to find the third term. To do that, you plug in 3 everytime there is an "n" in the formula.
*So you get t3=t3-1 + t3-2
*And that gives you t3=t2 + t1 which (using the numbers they gave you) is t3 = 4+2
*So t3 is 6.
*Now you want to find the 4th term. And to do that you plug in 4 everytime there's an "n" in the formula.
*So you get t4=t4-1 + t4-2
*So that's t4=t3 + t2 which is 6+4
*t4 is 10
*Last you need to find the 5th term. And you just plug in 5 everytime there's an "n" in the formula.
*So you get t5 = t5-1 + t5-2 which is t4 + t2
*And that gives you 10 + 6
*t5 = 16

*Now for what I don't understand. I'm having a little trouble with recursive definitions. I don't really get what you're supposed to do to find them. But that's about it, I get everything else. So if anyone would like to help me with that, that would be great (:

Reflection

So this week we started learning Chapter 13, it seems easy but i'm having trouble grasping the concept of this chapter. If you memorize the formulas, it helps a lot.
I'll explain some stuff from 13-2 and 13-3.

Recursive Definitions-formula for a sequence that involves the previous term (an-1) of tn-1

Example:

Find the recursive definition of 81, 27, 9, 3
an=an-1/3

Find the recursive definition of 1, 2, 6, 24, 120, 720
n=1 1
n=2 2 an-1Xn
n=3 6
n=4 24

13-3

Arithmetic Sn=n(t1+tn)/2
Geometric Sn=t1(1-r^n)/1-r

Example:

Find the sum of the first 10 terms of the series.
2-6+18-54+... geometric
r=-6/2=-3
S10=2(1-(-3)^10)/1-(-3)= -29,524

I'm kind of having trouble with this chapter so if anyone wants to explain it better, that would be great!