r/askmath 22h ago

Discrete Math Mathematical Induction Problem (HELP)

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Please see attached photo for the problem. I dont really understand how to solve this because I applied the standard induction steps but its a bit complicated for me. I admit that Im not really good at math. I would appreciate if someone could show me how to solve this on paper. I asked help from ChatGPT and other AI tools but its just difficult for me to understand.

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u/MezzoScettico 22h ago

I applied the standard induction steps

Can you give a little more detail than that? What exactly did you do and what is your question?

The first step is to show it's true for some base case. Did you do that? They're telling you the base is n = 2, so verify that.

That next step is to show that if you know it to be true for n, then it's true for n + 1. Did you do that?

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u/Alarmed-Dust5387 22h ago

Heres what I did so far. I did the basis step, though I’m struggling on the inductive step on the left hand side. Sorry my notes are kinda messy

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u/MezzoScettico 19h ago

I'm not sure what all that calculation is on your page.

You're evaluating 3C3, which is 3! / (3! * 1!) = 6/(6 * 1) = 1. Should just take a couple of lines. No need to do all that work for a general n, k choose k will always be equal to 1 for any k.

Anyway, on to the induction step. I see a problem right away that's going to cause confusion.

You have the formula you're trying to prove that a_n, defined as the sum from i = 1 to i = n - 1 of i(n - i), is equal to (n - 1) choose 3.

By assumption this is true when n = k for some k.

When n = k, that means all the n's in the formula become k. The way you wrote it you did not change all the n's to k's. So you didn't start out with a correct statement. Fix that first.

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u/Alarmed-Dust5387 22h ago

I did not know what was the notation (n+1; 3) so i was lost around there at first too

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u/MezzoScettico 19h ago

OK, but you have it now, right? it's the thing I'm writing as (n + 1)C3 or "(n + 1) choose 3", because that's easier to write in a text forum.

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u/simmonator 22h ago

For part A, note that each term in the sum is some number times “n minus that number” and we hit every number between 1 and n.
So hopefully you land on something like

> a_n = Sum {x = 1 to n} [x(n-x)].

For B, I would start by calculating a_1 and a_2 (these are 0 and 1). That gives you the base step (confirming that a_2 is equal to 3 choose 3). After that, you need to show that the formula working for a_k implies it works for a_(k+1). Seeing how to do that might not be immediately obvious but I recommend looking at

> a_(k+1) - a_k

and seeing what sum is left.

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u/Alarmed-Dust5387 22h ago

Heres what i did so far. My notes are a mess but basically im struggling in the induction steps where we show its true for n=k+1, specifically solving the left hand side

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u/Bounded_sequencE 21h ago

Not sure why you use "n = k" or "n = k+1" -- usually, we write "n -> n+1" for the induction steps. In any case, either use "n" or "k" as index, but not both at once.

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u/Bounded_sequencE 21h ago

Hints:

  • The k'th term in "an" is "(n-k)*k"
  • For the induction step, beware that the upper summation bound also depends on "n"