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

this it?

\(\large 1^2 + 4^2 + 7^2 + ... + (3n - 2)2 = \frac{n(6n^2-3n-1)}{2}\)

start by plugging in some values for n. that's how you'd start the induction proof, so go with n = 1, 2 etc until you make it work...or prove it wrong

Answer 2

I've done this I plugged n=1 and I got 1 for both. When I sub in n+1 for n that is when I ran into trouble If you could help me to understand that part I'd appreciate it.

Answer 3

now assume it's true for n = k so

\(\large 1^2 + 2^2 + ... (3k−2)^2=\frac{k(6k^2−3k−1)}{2}\) is true , OK?

and it follows by adding the same amount to LHS and RHS that

\(\large 1^2 + 2^2 + ... (3k−2)^2 + (3(k+1)−2)^2 = \frac{k(6k^2−3k−1)}{2}+ (3(k+1)−2)^2\)

i would next expand out the RHS. remember we want it to equal \(\large \frac{(k+1)(6(k+1)^2−3(k+1)−1)}{2}\) to prove the statement, so then long divide by k+1 and refactor what's left to get \(\large \frac{(6(k+1)^2−3(k+1)−1)}{2}\) or expand out \(\large \frac{(6(k+1)^2−3(k+1)−1)}{2}\) to get what's left, so that LHS = RHS

a lot of donkey work to be done

Answer 4

I ended up with 9n^2+3n+1=(6n^3+9n^2+5n+2)/2
Those two don't appear to be equal unless there's something I'm missing

Answer 5

pure unbridled joyless drudgery so let's hope i don't get my retricekicked for being "too helpful"

it is still a bit cryptic so i think i should be OK