A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 years ago
Use mathematical induction to prove that:
1/1*2+1/2*3+1/3*4+...+1/n*(n+1)=n/n+1
Have no idea where to start!
anonymous
 5 years ago
Use mathematical induction to prove that: 1/1*2+1/2*3+1/3*4+...+1/n*(n+1)=n/n+1 Have no idea where to start!

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well... one would think at the base case :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its obviously having to prove it for n>1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so check n=1 , its true , thats the first step done

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then assume it is true for n=k , ie that S(k) { the sum of k terms } = k / (k+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we need to prove it is true for n=k+1 , ie that S (k+1) = (k+1)/ ( k+2) where we got the RHS , by substituting n=k+1 in the summation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now there is a result that S(k+1) = S(k) + T(k+1) that is , in words, "the sum of k+1 terms is equal to sum of the first k terms , plus the (k+1)th term "

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like saying, sum 1 + 2 + 3 + 4 , well thats equal to summing the first 3 ( to get 6 ) and then adding on 4 ( the (k+1)th term )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now , back to the question , S(k+1) = k / (k+1) + 1/ (k+1)(k+2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0S(k+1) = [ k/(k+1) ] + 1/ [ (k+1)(k+2)] where the first fraction came from our assumption , that is what S(k) was assumed to be ! and the second term comes from subing n=k+1 into the general term on the LHS of the summation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now if you get S(k+1) over a single denominator and simplify , you will get S(k+1) = [ (k+1)/(k+2)] , which is what we set out to prove from step 3. Hence it is proved by induction the end

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, my! thank you! I would never have gotten this. Thank you for explaining it step by step. SOOOOOO helpful!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.