## anonymous one year ago 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

1. nuttyliaczar

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

2. anonymous

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

3. nuttyliaczar

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

4. nuttyliaczar

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

5. anonymous

At that point what would I do next though?

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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.

14. anonymous

Oh alright that makes sense

15. anonymous

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.

16. anonymous

$\frac{ k(6k^{2} - 3k - 1) }{ 2 } + [3(k+1)-2]^{2} = \frac{ 6k^{3} - 3k^{2} - k + 2(3k+1)^{2} }{ 2 }$ $= \frac{ 6k^{3} - 3k^{2} - k + 2(9k^{2} + 6k + 1) }{ 2 }$ $=\frac{ 6k^{3} - 3k^{2} - k + 18k^{2}+12k + 2}{ 2 } = \frac{ 6k^{3} +15k^{2}+11k+2 }{ 2 }$ Now here's where the tricky part comes in.

17. anonymous

Ill let you finish first

18. anonymous

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

19. anonymous

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.

20. anonymous

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

21. anonymous

So just a quick synthetic division: |dw:1433707447971:dw| So this means $\frac{ 6k^{3} + 15k^{2} + 11k + 2 }{ 2 } = \frac{ (k+1)(6k^{2}+9k+2) }{ 2 }$

22. anonymous

Bleh, lost what I was typing up, so gotta retype it, hang on

23. anonymous

no worries thank you so much!

24. anonymous

So, we need to get 6k^2 + 9k + 2 to look like 6(k+1)^2 -3(k+1) -1 If we were to expand that (don't actually expand this in your work, we don't want to mess with the right hand side, this is just to show you what we need to do) we would get 6k^2 + 12k + 6 -3k -3 -1 Conveniently, this is equivalent to what we have, so this is good. So let's just manipulate our left hand side to get this and then we can finish up $\frac{ (k+1)(6k^{2}+9k +2) }{ 2 } = \frac{ (k+1)(6k^{2}+12k-3k + 6-3-1) }{ 2 }$ $= \frac{ (k+1)(6k^{2}+12k+6 -3k-3 -1) }{ 2 }$ $= \frac{ (k+1)[6(k^{2}+2k+1)-3(k+1)-1] }{ 2 }$ $=\frac{ (k+1)[6(k+1)^{2} -3(k+1)-1] }{ 2 }$ So we got what we wanted. So you can conclude with something like: Therefore P(k+1) is true. Thus by the principal of mathematical induction, the statement P(n): 1^2 + 4^2 + 7^2 + ... + (3n-2)^2 = n(6n^2-3n-1)/n is true for all positive integers n. I feel like I butchered that a little, but honestly the tricky part is the algebra. I think the biggest trick to that was thinking to divide knowing that k+1 had to be a factor of what I was trying to simplify.

25. anonymous

Ive been trying to figure this out for a while now thanks a lot for your help!

26. anonymous

You're welcome