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

For n = 1 I got \[1^2 = 1\] after simplifying but do not know where to go from there.

Answer 2

next assume that the statement is true for n = k
and write the formula for n = k
Then find a formula for the (k + 1)th term in the series and add it to the statement for n = k
Then try to manipulate yhe latter to see if you can get it in the same form as for n = k
with the k replaced by k+1. If you can do this then you have proved the statement

Answer 3

I'm confused as to what the k is representing. I know to replace n with k but I don't understand what that mean.s

Answer 4

sum for n = k is k(6k^2 - 3k -1)/2

k represents any integer value of n

Answer 5

(k + 1)th term = 3(k + 1) - 2)^2

Answer 6

Oh okay, so for n = k to prove it is all I have to do is replace n with k? Are there any other steps? And with K+1 is it the same thing, just replacing n with k+1?

Answer 7

You add the (k+1) term to the formula for k:-

Sum of k+1 terms = n(6n^2 - 3n -1)/2 + (3k - 1)^2

Answer 8

Sorry that should be
k(6k^2 - 3k -1)/2 + (3k - 1)^2

Answer 9

Oh okay, that makes sense. Are there any other steps to prove it or is it already proved?

Answer 10

no This is the hard part. You need to convert this to 'the formula for k' replaced by k+1

in other words something like

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

- do you see the pattern - the 'k' in original formula is rplaced by k+1

Answer 11

What I have so far:
For n = 1
\[1^2 = 1(6 \times 1^2 - 3 \times 1 - 1)/2 = 1\]
For n = k
\[k(6k^2 -3k - 1)/2\]
For k+1
\[(3(k+1)-2)^2 + k(6k^2-3k-1)/2\]

Answer 12

yes ths correct so far

Answer 13

Okay I see that the k was replaced by k+1, I don't know what to do next though.

Answer 14

no its a tricky on because the trinomial won;t factor
I'll have a look at it..

Answer 15

In the directions it says the statement can be false, I don't know if that helps.

Answer 16

Well thats what i think . I cant see how we can manipulate that to get the required expression.

Answer 17

The only problem is I have only two questions and the other one is already false so I'm doubtful they would give me two false questions.

Answer 18

Lets ee if its tru for n = 3
3(6*9 - 9 - 1) / 2 = 66
yes its true for n = 3

thats not a proof though

Answer 19

So it is true we just need to figure out how to simplify it to get it the n=k statement?

Answer 20

if its true we will get the sum of k statement with the k replaced by k+1.
But the algebra is a bit hairy!

Answer 21

Okay, I think I'll work on it for a bit. Thank you for all your help.

Answer 22

I guess you could simplify your expression to a cubic polynomial.
Then you do the same with the original expressiion with the n replaced by k + 1.
If these 2 agree then it is proved.

Answer 23

yes - I got a result that way

But i cant tell you if its true or false.