Hard mathematical induction question

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Discussion Overview

The discussion revolves around a mathematical induction problem, specifically proving a statement involving products of integers and powers of two for all integers n greater than or equal to 1. Participants explore the steps of the proof and clarify their understanding of the induction process.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Exploratory

Main Points Raised

  • One participant presents the induction problem and the initial steps of the proof, expressing confusion about the assumption made for n = k.
  • Another participant clarifies the assumption and suggests that the left-hand side (LHS) of the statement for n = k + 1 relates to the LHS for n = k divided by (k + 1).
  • There is a discussion about the form of the LHS and the variables involved, with a focus on understanding the structure of the induction step.
  • A later reply encourages exploration and experimentation with the statements being proved, emphasizing that the process of arriving at proofs can be messy and non-linear.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the induction assumption, but there is a shared understanding of the need to explore the proof process more deeply.

Contextual Notes

Some participants express uncertainty about the notation and the specific powers involved in the induction steps, indicating potential limitations in the clarity of the original problem statement.

Who May Find This Useful

This discussion may be useful for students learning about mathematical induction, particularly those struggling with the assumptions and steps involved in proofs.

ZOMGbirdy
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Hi everyone, I have been trying to do this induction question... but can't seem to understand how to do it.

Prove by induction that, for all integers n is greater than or equal to 1,

(n+1)(n+2)...(2n-1)2n = 2^n[1x3x...x(2n-1)]

This is what someone else has given as the answer:

For n = 1
LHS = (1 + 1) = 2
RHS = 21 = 2
LHS=RHS hence statement is true for n = 1
Assume the statement is true for n = k
(k + 1)(k + 2)...(2k - 1)2k = 2k[1 x 3 x ... x (2k - 1)]
Required to prove statement is true for n = k + 1
(k + 2)(k + 3)...(2k + 1)(2k + 2) = 2k + 1[1 x 3 x ... x (2k + 1)]
LHS = (k + 2)(k + 3)...(2k + 1)(2k + 2)
= 2(k + 1)(k + 2)(k + 3)...2k(2k + 1)
= 2(2k + 1)2k[1 x 3 x ... x (2k - 1)] by assumption
= 2k + 1[1 x 3 x ... x (2k - 1) x (2k + 1)]
= RHS
If the statement is true for n = k, it is true for n = k + 1
Since the statement is true for n = 1, it is true for all positive integers n by induction.

I think I understand everything up until he actually makes the assumption. I don't understand the assumption and what he is actually subbing into the assumption.
 
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(k+1)(k+2)(k+3)...(2k-1)2k = 2k[1x3x5...x(2k-1)] = P(k)
... assumed true. (note: you left off the power?)

(k+2)(k+3)(k+4)...(2k+1)2(k+1) = 2(k+1)[1x3x5x...x(2k+1)] = P(k+1)
... needs to be proved. Kinda suggestive isn't it?

Consider:
Is the LHS of P(k+1) the same as the LHS of P(k)/(k+1) ?

The form of the LHS of P(n) is
f1f2f3...fm=Pm(k)

where
fi = k+i

but for i=m, fm = k+m = 2k(2k-1)
so what is fm-1 ?
 
Sorry, didn't realize the powers weren't copied properly.

Thanks for your reply, yea I get it now - I was looking at it from the wrong perspective.
 
No worries - when you are shown example proofs, they are all so very polished and clever. What you should realize is that these are the finished proofs - that is not how they were arrived at.

Don't be frightened to play around with the statements you are asked to prove. Mess about with them - figure out how they work. Look for patterns. Ask what these statements are telling you. This part is quite untidy and inelegant but you'll learn more that way.

Have fun :)
 

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