Question

A statement Sn about the positive integers is given. Write statements S1, S2, and S3, and show that each of these statements is true. Show your work.
Sn:   12 + 42 + 72 + . . . + (3n - 2)2 = n(6n^2-3n-1)/2

Answer 1

Sn:   1^2 + 4^2 + 7^2 + . . . + (3n - 2)2 = n(6n^2-3n-1)/2

Answer 2

@Agl202 any thoughts?

Answer 3

first off can someone explain what the +...+

Answer 4

So you want to show \[\sum_{i=1}^n(3i-2)^2=\frac{n(6n^2-3n-1)}{2}\]

Answer 5

The \(+..+\) means it keeps going on in this way...

Answer 6

where did you get the \[\sum_{i=1}^{n}\]

Answer 7

@zzr0ck3r

Answer 8

This is just what it means to add up a bunch of things.

Answer 9

Do you know what induction is?

Answer 10

ok and no please explain

Answer 11

What class is this for, and how is it that you have a question that you seemingly know nothing about?

Answer 12

pre calculus and because i have no teacher and am self teaching this

Answer 13

Well the only way I know how to show this is through induction and I don't have time to teach you how to do all of that... Go read on induction and then you should be able to do this

Answer 14

This seems like a little overboard for pre calc.

Answer 15

could you just tell me what i need to do to solve it like what am i trying to get to

Answer 16

Well \(\sum_{i=1}^n(3i-2)^2=1^2+4^2+7^2+10^2+13^2+...+(3n-2)^2\)

We want to show \(\sum_{i=1}^n(3i-2)^2=\frac{n(6n^2-3n-1)}{2}\) for every natural number.

To prove such things about the natural number, we use indiction. Induction says that if you can show something is true for \(n=1\) and you can show its true for \(n+1\) assuming it is true for \(n\) then it is true for all \(n\).

For sure \(\sum_{i=1}^n(3i-2)^2=\frac{n(6n^2-3n-1)}{2}\) when \(n=1\) because \(\sum_{i=1}^1(3i-2)^2=1\) and \(\frac{1(6*1^2-3*1-1)}{2}=1\).

Now we assume it's true for some \(n\) and want to show its true for \(n+1\).

\(\sum_{i=1}^{n+1}(3i-2)^2= (3(n+1)-2)^2+\sum_{i=1}^n(3i-2)^2 = (3n+1)^2+\frac{n(6n^2-3n-1)}{2}=\\ \frac{(n+1)(6(n+1)^2-3(n+1)-1)}{2}\) and thus it is true for all \(n\).

Answer 17

thank you

Answer 18

np

Answer 19

@zzr0ck3r one last thing i want to ask about this?

Answer 20

yes?

Answer 21

how does s1, s2, and s3 come into play

Answer 22

This is a big question, google partial sums.

http://www.mathsisfun.com/algebra/partial-sums.html

Answer 23

@zzr0ck3r ok i understand it now i think so what does s1 stand for than?

Answer 24

There is simply no way you read that...

S1 is the first term
S2 is the sum of the first 2 terms
S3 is the sum of the first 3 terms
.
.
.
Sn is the sum of the first n terms

Answer 25

so now im confused if sn is the statement i work with how do i write statment 1-3

Answer 26

like the sum of what 2 terms for s2 for instance

Answer 27

S2 = 1^2 + 4^2
S3 = 1^2+4^2+7^2
Sn = 1^2+4^2+7^2+10^2 + ... + (3n-2)^2

Answer 28

thats what i thought

Answer 29

note that 3(1) - 2 = 1
3(2) - 2 = 4
3(3)-2 = 7
3(4) - 2 = 10
.
.
.

Answer 30

so since s2 is 1^2+4^2 would i plug nine into n for s2?

Answer 31

im confused on what to do with the equation in connection to those terms

Answer 32

ok so lets look at S3

1^2+4^2+7^2 = 1+16+49 = 66

now let \(n=3\) in \(\frac{n(6n^2-3n-1)}{n}\) and we get \(\frac{3(6*3^2-3*3-1)}{2}=66\)

Answer 33

So this will always work for any n you choose.

Answer 34

ohhhh

Answer 35

so its finding n so it equals the sum of the three terms or how ever many we use!!!

Answer 36

no, we dont find n, it works for all n

Answer 37

but you would not want to show that it works for n = 23450987565746553738

because that would take your entire life, but since I proved it with a technique called induction we know its true for all n, including n = 23450987565746553738

Answer 38

no i mean for s1 s2 and s3 its pluggin in a number for n that work for the sum of those terms used which intern says that all n works if it works for those

Answer 39

your using sn to prove s1 s2 and s3 are true for this particular question

Answer 40

no, that does not show it works for all n. The thing I did earlier shows that it works

induction is basically this. if you can show that when something is true for n it must be true for n+1, and you can show its true for 1, then it must be true for 2 and then in turn it must be true for 3 and so on.

So induction you do three things

1) show its true for n=1
2) assume its true for some arbitrary n
3) show that it must follow that it is true for n+1

Answer 41

its like dominoes, if the one before you falls, then you are going to fall

Answer 42

and if you start on the first domino, they will all fall

Answer 43

thats what i mean though, is this question asking to prove sn to be true or for you to right the statements for term1 1 and than term 1+ term 2 and than term 1+ term 2 +term 3 showing what you must put into n to equal those sum of those terms

Answer 44

im confused i think

Answer 45

the question is asking you to show that for any \(n\) you choose the statement holds.

Like I chose 3 and showed it held, and I could pick 4 and show it holds, and I could pick 27 and show it holds, but no matter how many cases I show it holds for does not prove that it holds for all cases. For that we need induction

Answer 46

but what i dont get is for each statment i write how do i know what number to use for n

Answer 47

I dont understand what you mean

Answer 48

i have to write statement 1 statement 2 and statement 3. than show the statments are true so for instance
s1
1^2= 1
so how do you know what to plug into for n so that
(3n - 2)2 = n(6n^2-3n-1)/2 makes out to 1=1

Answer 49

i mean thats easy because its obviously 1 that you plug into it

Answer 50

but there other may not be

Answer 51

this is not what you are asked to do and I don't know any better way to explain it.

1^2+4^2+7^2+10^2+......+(3n-2)^2 will always equal n(6n^2-3n-1)/2 no matter what n you pick.

Answer 52

but i thought to write statements s1, s2 and s3 is just writing
1^2
1^2+4^2
1^2+4^2+7^2
and than prove that each of those is true

Answer 53

and i had assumed to prove individually that those are true you had to plug into n to make the equation equal the sum of those

Answer 54

yes and if you try and do that for every number in the world you are going to be here for an infinite amount of time

Answer 55

ohh so i just write s1 s2 and s3 than state that using induction proves that those statments can be found in (3n - 2)2 = n(6n^2-3n-1)/2

Answer 56

(3n - 2)2 = n(6n^2-3n-1)/2 only for n=1

but 1^2+4^2 = 2(6*2^2-3*2-1)/2

and

1^2+4^2+7^2 = 3(6*3^2-3*3-1)/2

and

1^2+4^2+7^2+10^2+14^2 +18^2 = 6(6*6^2-3*6-1)/2

Answer 57

but thats what i mean how do you find that plugging in 2 into the n(6n^2-3n-1)/2 will equate to the sum of those two terms

Answer 58

and the same for 3 instead of 2

Answer 59

thats where im hung up on

Answer 60

i mean what if you needed 67 instead of 2 how would you figure out you need 67

Answer 61

\(\sum_{i=1}^3(3i-2)^2=1^2+4^2+7^2=\frac{3(6*3^2-3*3-1)}{2}\\ \sum_{i=1}^4(3i-2)^2=1^2+4^2+7^2+10^2=\frac{4(6*4^2-3*4-1)}{2}\\ \sum_{i=1}^5(3i-2)^2=1^2+4^2+7^2+10^2+14^2=\frac{5(6*5^2-3*5-1)}{2}\\

\sum_{i=1}^n(3i-2)^2=1^2+4^2+7^2+...+(3n-2)^2=\frac{n(6*n^2-3*n-1)}{2}\)

Answer 62

you want to prove the last line

Answer 63

ok good luck, I have to go. Just so you know.

I did not learn anything about series (Sn) until calc 3, and I was not asked to prove things until 300 level math classes. I think you might be self studying things that you might not use for up to a year after you start calc, and thus are wasting your time.

Answer 64

this is a pre calc class im studying for :(

Answer 65

its part of my highschool curriculum