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 = quantity n times quantity six n squared minus three n minus one all divided by two

Answer 1

I suggest writing out that second half of the equation in terms like the first. ...=n(6n^2-3n-1)/2

Answer 2

I know the statement for n=1 is true but I don't know how to find if k+1 is true

Answer 3

Just check the equation again using the next integer, which would be 2

Answer 4

Oh true true I forgot how you're supposed to prove by induction. Yes @Concentrationalizing is right to substitute all the n's with n+1 for the next part

Answer 5

At that point what would I do next though?

Answer 6

Okay, so we want to show this:
\[1^{2} + 4^{2} + 7^{2} + ... + (3k-2)^{2} + [3(k+1)-2]^{2} = \frac{ (k+1)[6(k+1)^{2} - 3(k+1) - 1] }{ 2 }\]

Answer 7

Sorry that it got cut off, its just a -1 in the numerator that got cut off.

Answer 8

So by the induction hypothesis, we assumed this to be true:
\[1^{2} + 4^{2} + 7^{2} + ... + (3k-2)^{2} = \frac{ k(6k^{2}-3k-1) }{ 2 }\]

Answer 9

Thus we can do this substitution:
\[\frac{ k(6k^{2} -3k-1) }{ 2 } + [3(k+1)-2]^{2} \]

Answer 10

Now just to do the algebra and try to get it to look like k+1

Answer 11

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

Answer 12

why do you add on the + [3(k+1)-2]^2

Answer 13

Well, on the left side we have an infinite sum. Every consecutive term on the left must be added. When we do the k+1 term, its equivalent to replacing the k in (3k-2)^2 with k+1. But since its an infinite sum, that k+1 term must be added.

Our original assumption though was that 1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k(6k^2-3k-1)/2 and you cannot simply add the [3(k+1)-2]^2 unles its added to both sides.

Answer 14

Oh alright that makes sense

Answer 15

Mhm. So the first thing I did was use our induction hypothesis to make the substitution on the left hand side. Now Im just trying to combine and simplify to make it look like the right.

Answer 16

\[\frac{ k(6k^{2} - 3k - 1) }{ 2 } + [3(k+1)-2]^{2} = \frac{ 6k^{3} - 3k^{2} - k + 2(3k+1)^{2} }{ 2 }\]
\[= \frac{ 6k^{3} - 3k^{2} - k + 2(9k^{2} + 6k + 1) }{ 2 }\]
\[=\frac{ 6k^{3} - 3k^{2} - k + 18k^{2}+12k + 2}{ 2 } = \frac{ 6k^{3} +15k^{2}+11k+2 }{ 2 }\]

Now here's where the tricky part comes in.

Answer 17

Ill let you finish first

Answer 18

so after the algebra my answer would be
1^2+4^2+7^2+... +(3k-2)^2+[3(k+1)-2]^2=(k+1)[6(k+1)^2-3(k+1)-1]/2

Answer 19

Thats what youre hoping to get to. We need to manipulate the left hand side to make it look like the right, though. I have the left simplified down to what I have above.

Answer 20

Where I have it right now is where it gets tricky, but here's what we can do. Now, notice the right hand side has a (k+1) multiple. Basically this means that k+1 is a factor of [6(k+1)^2 -3(k+1)-1], otherwise theres no way itd be factored out as it is. So what I want to do is take the numerator I have and divide it by k+1 and see what the remainder is

Answer 21

So just a quick synthetic division:


So this means
\[\frac{ 6k^{3} + 15k^{2} + 11k + 2 }{ 2 } = \frac{ (k+1)(6k^{2}+9k+2) }{ 2 }\]