anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
IrishBoy123
  • IrishBoy123
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
anonymous
  • anonymous
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.
IrishBoy123
  • IrishBoy123
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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
IrishBoy123
  • IrishBoy123
1 Attachment
IrishBoy123
  • IrishBoy123
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.