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For my theory of computation class, we are supposed to do some review/practice problems to work off the rust and make sure we are ready for the course. Some of the problems are induction proofs. I did this at one time, but apparently it has completely escaped me. I've watched a couple tutorials, but still can't do problem 'a'. If anyone can walk me through the first problem I'm pretty sure I could figure out the second one on my own. Any help would be appreciated!

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Kara
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Bryan
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    This is more appropriate for [math.stackexchange](http://math.stackexchange.com). – arshajii Sep 10 '13 at 22:38
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    This question appears to be off-topic because it is about math. – millimoose Sep 10 '13 at 22:47
  • The basic idea for problem a) is that you take the expression and substitute `n+1` in place of `n`. You then fiddle with the expression until you get it into the form [`n(n+1)(2n+1)/6 + (n+1)^2`](http://goo.gl/CoZVgT). This is equal to [`(sum i = 1 to n of i^2) + (n+1)^2`](http://goo.gl/i3zx5R), which is equal to [`sum i = 1 to (n+1) of i^2`](http://goo.gl/Rpyqx7), proving the induction step holds. – millimoose Sep 10 '13 at 22:55
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    The general idea is that you have to somehow express the value of a term for `n+1` using the value of the term `n`, and the obvious step between these steps as the appropriate function of `n+1`. (E.g. for a sum, the next step is adding `n+1` to the previous term. For a factorial, it's multiplying. For `2^n`, it's multiplication by 2.) Now, a snag in the works is that the approach for a) doesn't really seem to neatly transfer to an inequality such as is in b). – millimoose Sep 10 '13 at 23:02
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    That said, you might be able to show that for all steps greater than a chosen starting value of `n`, the left side is multiplied by a greater number than the right side, which is always multiplied by `2`. (This is in fact pretty obvious for all `n` greater than 2, but you might have to be a little formal.) – millimoose Sep 10 '13 at 23:04
  • Thanks for the help and I wasn't aware of the math.stackexcahnge. Sorry for the offtopic post. – Bryan Sep 10 '13 at 23:06

1 Answers1

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First verify it holds for n = 1.

Then assume it is true for n = x ( the sum of the first x squares ) and then try to compute the sum of the the first x + 1 squares. You know the result for the first x, you just add the last square to that sum. From there it should be easy.

And you posted on the wrong site.

tomato
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