text: # Inductive proof of summation formula
  
  livcon
  
  ## Homework Statement
  
  Prove by induction the following summation formula:
  $$\frac{1}{1\1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
  $$n \geq 1$$
  
  -
  
  ## The Attempt at a Solution
  
  Inductive step:
  1. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{(n+1)+1}$$
  
  2. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$
  
  To prove this (1) should equal (2) (right?), but I don't see how to manipulate (2) to make it equal to (1)
  
  whs
  Common denominator!
  
  Staff Emeritus
  Gold Member
  but I don't see how to manipulate (2) to make it equal to (1)
  You don't need to. All you need to do is to show these two sums of fractions are equal.
  
  livcon
  That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks
  
  Staff Emeritus
  Gold Member
  As a warmup exercise, prove the following:
  
  $$\frac{1}{3} + \frac{1}{6} = \frac{3}{2} - 1$$
  
  emyt
  That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks
  
  see how 1 - 1/n+1 looks first
  
  livcon
  I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.
  
  Staff Emeritus
  Gold Member
  I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.
  Isn't proving
  $$1 - \frac{1}{(n+1)+1} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$​
  exactly the same kind of problem, though? Why can't you do the same thing you did for the warmup problem?
  
  livcon
  I guess it's because in the warmup exercise I could easily add and subtract the fractions because it was all constants, but in the one involving the variable n I got stuck. So I would probably have an epiphany if someone would just write it out. This is just a matter of me not seeing the obvious steps in this particular problem.
  
  $$1- \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}$$
explanation: The document is a discussion thread from a homework help forum where a user is seeking assistance with an inductive proof of a summation formula. The user presents the problem and their attempt at a solution, and other users provide hints and guidance. The document contains a math question and an ongoing discussion about the solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Inductive proof of summation formula
  
  livcon
  
  ## Homework Statement
  
  Prove by induction the following summation formula:
  $$\frac{1}{1\1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}$$
  $$n \geq 1$$
  
  -
  
  ## The Attempt at a Solution
  
  Inductive step:
  1. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{(n+1)+1}$$
  
  2. $$\frac{1}{1*2} + \frac{1}{2*3} + ... + \frac{1}{n(n+1)} + \frac{1}{(n+1)((n+1)+1)} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$
  
  To prove this (1) should equal (2) (right?), but I don't see how to manipulate (2) to make it equal to (1)
  
  whs
  Common denominator!
  
  Staff Emeritus
  Gold Member
  but I don't see how to manipulate (2) to make it equal to (1)
  You don't need to. All you need to do is to show these two sums of fractions are equal.
  
  livcon
  That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks
  
  Staff Emeritus
  Gold Member
  As a warmup exercise, prove the following:
  
  $$\frac{1}{3} + \frac{1}{6} = \frac{3}{2} - 1$$
  
  emyt
  That's right, but in this case I've actually been staring at the problem for a while, so I'm really interested in how to do the algebra in (2). Thanks
  
  see how 1 - 1/n+1 looks first
  
  livcon
  I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.
  
  Staff Emeritus
  Gold Member
  I appreciate the help, but I'm afraid I'm still blind here, proving the equivalence in the warmup exercise is easy Hurkyl, but I just don't see it in my problem.
  Isn't proving
  $$1 - \frac{1}{(n+1)+1} = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}$$​
  exactly the same kind of problem, though? Why can't you do the same thing you did for the warmup problem?
  
  livcon
  I guess it's because in the warmup exercise I could easily add and subtract the fractions because it was all constants, but in the one involving the variable n I got stuck. So I would probably have an epiphany if someone would just write it out. This is just a matter of me not seeing the obvious steps in this particular problem.
  
  $$1- \frac{1}{n+1}+ \frac{1}{(n+1)(n+2)}$$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread from a homework help forum where a user is seeking assistance with an inductive proof of a summation formula. The user presents the problem and their attempt at a solution, and other users provide hints and guidance. The document contains a math question and an ongoing discussion about the solution.
  
  Label: question-and-answer