- anonymous

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(6𝑛2−3𝑛−1)2

- katieb

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- anonymous

n(6n^2-3n-1)/2

- anonymous

anybody got a clue

- anonymous

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## More answers

- phi

can you make this clearer
(3n - 2)2 ?

- anonymous

n(6n^2-3n-1)/2

- phi

No, I mean the last term in the sum
12 + 42 + 72 + . . . + (3n - 2)2

- anonymous

its 1^2+4^2+7^2

- anonymous

im so lost

- phi

oh. it looks like twelve + forty-two+...
and I was not making any progress (who would have guessed)

- anonymous

lol its my fualt i should have double checked thst sooo

- phi

Do you have to prove the result or derive it?
proving it should be easier

- anonymous

no i just have to show thast the stamnets are true and show all my work proving them

- phi

are you studying proof by induction?

- anonymous

i think so

- anonymous

yes iam

- phi

the first step is show the identity is true for n=1
can you do that ?

- anonymous

no im jsut now learning this stug=ff lol sorry im a newbie

- phi

when n=1, what is "series" ?

- phi

what is the last term of the summation when n=1 ?

- anonymous

1

- phi

and the starting term is 1
in other words the entire summation is
1^2
(just 1 term)
on the right-hand side we have the formula
n(6n^2-3n-1)/2
what is that when n=1 ?

- anonymous

ummmmm

- phi

you sub in n=1 into the formula n(6n^2-3n-1)/2
what do you get ?

- anonymous

scratches head um n+1(6n^2-3n-1)/2

- phi

the idea is everywhere you see n in the formula, erase it, and write in 1 in its place
then simplify (now that you have a number instead of a letter, you can do that)
what do you get ?

- anonymous

1(61^2-31-1)/2

- phi

ok. and remember when you have for example 3n that means 3*n
( multiply signs are usually left out)
so the formula is
1*(6*1^2 - 3*1 -1)/2
now simplify that

- anonymous

hold on im doing it now

- anonymous

the answer is 1 right

- phi

yes, it simplifies to (6-4)/2 = 2/2 = 1
that shows the formula works for n=1
because
1^2 = 1 (the answer we get from the formula)

- phi

now the hard part.
we assume the formula works for numbers up to n (whatever n happens to be)
and show it also works for n+1
This part is difficult because it looks like we need some messy algebra.

- anonymous

this is what i didi
1^2 evaluates to 1
Multiply 1 and 6
1
6*1^2 evaluates to 6
Multiply 1 and 3
1
3*1 evaluates to 3
6*1^2-3*1 evaluates to 3
6*1^2-3*1-1 evaluates to 2
Multiply 1 and 2
1
1*(6*1^2-3*1-1) evaluates to 2
1*(6*1^2-3*1-1)/2 evaluates to 1

- anonymous

uhhh math

- anonymous

so how dowe start this

- phi

yes, that is how you evaluate it.

- anonymous

sweet

- phi

all of that proves that the formula works for n=1
now assume the formula works for numbers up to n
show it works for n+1
Induction Step:
Assume the identity is true for n:
\[ 1^2 + 4^2 + ... +(3n-1)^2 = \frac{n(6n^2-3n-1)}{2} \]

- phi

now add the next term to the summation
that is add \( \left(3(n+1)-1 \right)^2 \) to both sides
\[ 1^2 + 4^2 + ... +(3n-1)^2+\left(3(n+1)-1 \right)^2 = \frac{n(6n^2-3n-1)}{2} +\left(3(n+1)-1 \right)^2\]

- anonymous

holy moly tat is hard

- phi

yes, it is very difficult.

- anonymous

so is there more too it plz say no lol

- phi

the formula for n+1 becomes
\[ \frac{(n+1)}{2} \left(6(n+1)^2 -3(n+1) -1\right) \]
we have to show that
\[ \frac{n(6n^2-3n-1)}{2} +\left(3(n+1)-1 \right)^2 \]
can be written as \[ \frac{(n+1)}{2} \left(6(n+1)^2 -3(n+1) -1\right) \]

- anonymous

thnx

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