Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

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

Answer 1

@phi

Answer 2

First, prove the base case. Is the formula true for n=1 ?

Answer 3

Yes

Answer 4

so write that part down. base case is true.
now assume the formula is correct for n= k
is it true for k+1 ?

to do this part involves algebra

Answer 5

Yes, that's where I get stuck. I don't know how to do that.

Answer 6

on the right side, replace n with (k+1)

Answer 7

For proving for n=k+1 what you do is this. You've assumed n=k is correct so write what you got for assuming n=k down. Then add on 1 more term of (3n-2)^2 where n=k+1.

Answer 8

So would the new equation be (k+1)(6(k+1)^2-3(k+1)-1)/2?

Answer 9

on the left side, show the last "valid" term (replace n with k)
and add one more term, with n replaced with k+1

Answer 10

No. You need to put in the equation when n=k so k(6k^2-3k-1)/2 then add on one more term where n=k+1 so 3((k+1)-2)^2.

Answer 11

I am so lost.

Answer 12

Right, when you assume n=k you get.\[\sum_{r=1}^{k}(3r-2)^2 = k(6k^2 - 3k - 1)/2\]

Answer 13

For proving n=k+1 you put that Right hand equation back in and just add on one more term of (3r-2)^2 where r=k+1

Answer 14

So it would be \[k(6k^2 - 3k - 1)/2 + (3k+3-2)^2\]

Answer 15

And you want to prove that that is equal to n(6n^2-3n-1)/2 when n=k+1

Answer 16

(k+1)(6(k+1)^2-3(k+1)-1)/2. You're trying to prove it's equal to that

Answer 17

Okay, would I replace k with 1?

Answer 18

No. You need to expand the brackets and then factorize out the correct factors which will be k+1 and 1/2

Answer 19

How do I do that?

Answer 20

\[\frac{ 6k^3 - 3k^2 - k }{ 2 } + (3k +1)^2\]

Answer 21

Okay, then what?

Answer 22

Expand the (3k+1)^2

Answer 23

Do I FOIL (3k+1)(3k+1)?

Answer 24

Yes

Answer 25

9k^2 + 6k +2

Answer 26

+1 not +2

Answer 27

Ah, okay. Then what?

Answer 28

Factorize the 1/2

Answer 29

English please? Lol

Answer 30

take 1/2 out as a factor

Answer 31

Can I ask why you're doing induction?

Answer 32

Because I'm in pre-calculus...

Answer 33

But, I got 1/2(18k^2+12k+4)

Answer 34

+2

Answer 35

Yeah.

Answer 36

And then add in the first half that you've factorized also

Answer 37

1/2(6k^3-3k^2-k+18k^2+12k+2)

Answer 38

Then collect the terms.

Answer 39

1/2(6k^3+15k^2+11k+2)

Answer 40

Then factorize out k+1

Answer 41

now simplify the right side and show it equals that expression

Answer 42

This is a real b*tch to simplify which is annoying.

Answer 43

If I were you, I'd get someone who knows how to do proof by induction to go through it with you in person on paper because it is much easier like that than trying to explain with type.

Answer 44

(6k^3+15k^2+11k+2)/2 = ? (k+1)(6(k+1)^2-3(k+1)-1)/2
or
6k^3+15k^2+11k+2 = ? (k+1)(6(k+1)^2-3(k+1)-1)
you are trying to show the left side equals the right side.
perhaps the easiest way is to expand the right side, collect terms...

Answer 45

The left side came from the work you and alright just did
the right side came from putting k+1 into the formula


working on the right side
(k+1)(6(k+1)^2-3(k+1)-1)

concentrate on (k+1)^2 that is (by foil) k^2+2k+1
so we have
(k+1)( 6( k^2+2k+1) - 3(k+1) -1)
can you finish ?