How can you prove this discrete math induction statement?

AI Thread Summary
The discussion focuses on proving a discrete math induction statement involving the summation of the form n(n-1)(n+1). Participants emphasize the importance of establishing a base case and assuming the statement holds for n=m before proving it for n=m+1. They suggest reshaping the expression to match the desired form, specifically N*(N+1)*(N+2)/3, to demonstrate its validity for all N. A user recommends avoiding variable reuse for clarity and proposes breaking up the summation for the n=k+1 case as a helpful hint. The conversation highlights the structured approach required for mathematical induction proofs.
iPetey
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Homework Statement



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Homework Equations



base case: n=1

The Attempt at a Solution



im not sure where to start because the examples that my professor showed us did not have a n(n-1) (n+1) but rather (p+1)P=1+1)(2(p+1)+1)

im just very lost in this example
 
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Assume the statement is true for n=m. Add one more term and see if the statement is still true for n=m+1. If it is and also true for n=2, then true for all n...
 
you do this kind of stuff by induction
 
i know its done by induction but i don't know the proper steps to do it.
 
Assume that the statement you have is true. Now add another term (do the sum to N instead of N-1). Add that term to the RHS and see if you can reshape that expression to the same form as it is now, only with N*(N+1)*(N+2)/3 instead. If you can, it must hold true for all N from there on. Then it remains to show that it holds for a sum with just one term.
 
SEngstrom said:
Assume that the statement you have is true. Now add another term (do the sum to N instead of N-1). Add that term to the RHS and see if you can reshape that expression to the same form as it is now, only with N*(N+1)*(N+2)/3 instead. If you can, it must hold true for all N from there on. Then it remains to show that it holds for a sum with just one term.
Personally I don't like re-using the variable N; instead I would use k:
Assume true for n = k:
\displaystyle \sum_{i=1}^{k-1} i(i +1) = \frac{k(k-1)(k+1)}{3}
Prove true for n = k + 1:
\displaystyle \sum_{i=1}^{(k+1)-1} i(i +1) = ... = \frac{(k + 1)(k)(k+2)}{3}

OP: a hint would be to "break up" the summation in the n = k + 1 case.
 
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