Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

  • tux

Prove by induction: http://i.imgur.com/715S5.png

Mathematics
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Proof: For each positive integer \(n\ge2\) , let \(S(n)\) be the statement\[1.2+2.3+...+(n-1)n=\frac{n(n-1)(n+1)}{3}\] Basis step: S(2) is the statement \(1.2=\frac{2.1.3}{3}=2\). Thus \(S(2)\) is true. Inductive step: We suppose that \(S(k)\) is true and prove that \(S(k+1)\) is true. Thus, we assume that \[1.2+2.3+...+(k-1)k=\frac{k(k-1)(k+1)}{3}\] and prove that \[1.2+2.3+...+k(k+1)=\frac{k(k+1)(k+2)}{3}\] what to do now?
@tux make sense?
  • tux
You rewritten sum as \[\sum_{i=1}^{n-1}i(i+1)=\sum_{i=1}^{n}i(i+1)+(n-1)(n-1+1)\] ? And then substituted n=k in induction step Then n=k+1 you got ((k+1)-1)(k+1) k*(k+1)=\[\frac{ k(k+1)(k+2) }{ 3 }\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

this is how we got things done by induction We assume that \(S(k)\) is true and we want to prove that \(S(k+1)\) is true.
  • tux
Thank you. Now I can do it alone
yw :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question