sasogeek
  • sasogeek
prove by induction that \(\large 1 \times 2 + 2 \times 3 + ... +n(n+1) = \frac{1}{3}n(n+1)(n+2) \)
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mimi_x3
  • Mimi_x3
well, where are you stuck i assume you should know the steps n=k, then n=k+1
Mimi_x3
  • Mimi_x3
Assume \(n=k\) is true \[1*2+2*3+...+k(k+1) = \frac{1k}{3}(k+1)(k+2)\] Prove \(n=k+1\) \[1*2+2*3+...+k(k+1)+(k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}\] \[\frac{(k)(k+1)(k+2)}{3} + (k+1)(k+2) = \frac{(k+1)(k+2)(k+3)}{3}\] Now, you can prove RHS = LHS
sasogeek
  • sasogeek
ok i understand and have worked the up to the point \(\large 1* 2 + 1* 3 + ... + k+1(k+2) = 1 * 2 + 1*3 + ... +k(k+1) +k+1(k+2) \) \(\large = \frac{1}{3}k(k+1)(k+2)+(k+1)(k+2)\) then what?

Looking for something else?

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

More answers

RadEn
  • RadEn
for left side : k(k+1)(k+2)/3 + (k+1)(k+2) = (k+1)(k+2)(k/3 + 1) = (k+1)(k+2)(k+3)/3 same like right side
Mimi_x3
  • Mimi_x3
Well, you prove as I said above. Prove the LHS = RHS \[=>(k+1)(k+2)\left[ \frac{k}{3}+1\right]\]It should be straight forward now.
Mimi_x3
  • Mimi_x3
Just, some algebra and you're done!
sasogeek
  • sasogeek
it's still blurry but i'll try to get it in a bit :)
Mimi_x3
  • Mimi_x3
Well, where are you stuck?
Mimi_x3
  • Mimi_x3
All you have to do here is: Prove the LHS that is: \[\frac{(k)(k+1)(k+2)}{3} + (k+1)(k+2) \] Is equal to the RHS: \[=> \frac{(k+1)(k+2)(k+3)}{3}\]
sasogeek
  • sasogeek
ohh, thanks :) normal algebra takes off from there i see :) makes sense now xD

Looking for something else?

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