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(6𝑛2−3𝑛−1)2

Answer 1

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

Answer 2

anybody got a clue

Answer 3

@anonymous_user

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

im so lost

Answer 9

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

Answer 10

lol its my fualt i should have double checked thst sooo

Answer 11

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

Answer 12

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

Answer 13

are you studying proof by induction?

Answer 14

i think so

Answer 15

yes iam

Answer 16

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

Answer 17

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

Answer 18

when n=1, what is "series" ?

Answer 19

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

Answer 20

1

Answer 21

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 ?

Answer 22

ummmmm

Answer 23

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

Answer 24

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

Answer 25

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 ?

Answer 26

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

Answer 27

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

Answer 28

hold on im doing it now

Answer 29

the answer is 1 right

Answer 30

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)

Answer 31

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.

Answer 32

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

Answer 33

uhhh math

Answer 34

so how dowe start this

Answer 35

yes, that is how you evaluate it.

Answer 36

sweet

Answer 37

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:
\[ 1^2 + 4^2 + ... +(3n-1)^2 = \frac{n(6n^2-3n-1)}{2} \]

Answer 38

now add the next term to the summation
that is add \( \left(3(n+1)-1 \right)^2 \) to both sides

\[ 1^2 + 4^2 + ... +(3n-1)^2+\left(3(n+1)-1 \right)^2 = \frac{n(6n^2-3n-1)}{2} +\left(3(n+1)-1 \right)^2\]

Answer 39

holy moly tat is hard

Answer 40

yes, it is very difficult.

Answer 41

so is there more too it plz say no lol

Answer 42

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 \[ \frac{(n+1)}{2} \left(6(n+1)^2 -3(n+1) -1\right) \]

Answer 43

thnx