Help me with this exercise [Discrete mathematics]
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Can you show that the proposition is true for n = 1?Prove that View attachment 21751 for any integer n>=1 .
maybe it can be solved with mathematical induction... but i cant ..please help me
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Dr.Peterson
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Is that supposed to be a "t" before the sigma? If not, what is it?Prove that View attachment 21751 for any integer n>=1 .
maybe it can be solved with mathematical induction... but i cant ..please help me
We'll want to see how far you can get before you get stuck, so we can know what help will be most useful to you.
It does, however, look sufficiently different from the usual induction proofs you'd do in a discrete math class, that you might consider a different way. I believe it can be proved using the binomial theorem; in showing that it is true for n=1 I observed some interesting relationships that allow you to rewrite the LHS to look like the binomial theorem.
i mean i tried to start with n=1 but i dont know how to prove that both sides are equal .. here what i did so far
no theres nothing before sigma .
heres what i did but its nothing.. i dont know how to prove that both sides are equal so then i can go next step for n=k , n=k+1
Dr.Peterson
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Did you evaluate the left side?? It isn't a sum; it's a single term.
As I indicated, it seems hard to do the inductive step, so I looked at it differently. I observed, based on what the right side looks like after combining into a single fraction, that the left side could be written in a more interesting way by combining the first and last factors. In particular, if you add the term for k=0 to both sides, it can be written as the binomial theorem. I am assuming you have learned that.
i mean... i try to learn the solution for this exact exercise for my upcoming exam.. so it will be helpful if you try and saw me a proper solution.
Dr.Peterson
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No, you don't learn mathematics by memorizing "the solution for this exact exercise". There is no one correct method for doing a particular proof; and the main idea of a proof is the thinking that leads to discovering it. That is, you learn by thinking for yourself; I have seen no evidence that you have thought at all.
Can you at least confirm whether you have learned the binomial theorem, so that it would make sense that you would be expected to use it here? Or do the instructions for the problem explicitly state that it should be proved by induction? (You said "maybe".) If I were tutoring you in person, I would have your book in my hands looking for such information; here, I have to depend on you to provide it. Have you still not read our guidelines?
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Showing you a complete solution for you to memorize will not help you on your upcoming exam because you will not be asked this same question on the exam.
You learn math by doing, not by seeing.
In the first step
[MATH]n = 1 \implies \sum_{k=1}^n (-t)^k \dbinom{n}{n - k} (t + 1)^{-k} = \sum_{k=1}^1 (- t)^k \dbinom{1}{1 - k} * (t + 1)^{-k} =[/MATH]
[MATH](-t)^1 \dbinom{1}{0} * (t + 1)^{-1} = - t * \dfrac{1!}{0! * (1 - 0)!} * \dfrac{1}{t + 1} = WHAT?[/MATH]
And [MATH]n = 1 \implies (t + 1)^{-n} - 1 = (t + 1)^{-1} - 1 = \dfrac{1}{t + 1} - 1.[/MATH]
How would you prove those two are equal to each other?
Things equal to the same thing are equal to each other.
You learn math by doing, not by seeing.
In the first step
[MATH]n = 1 \implies \sum_{k=1}^n (-t)^k \dbinom{n}{n - k} (t + 1)^{-k} = \sum_{k=1}^1 (- t)^k \dbinom{1}{1 - k} * (t + 1)^{-k} =[/MATH]
[MATH](-t)^1 \dbinom{1}{0} * (t + 1)^{-1} = - t * \dfrac{1!}{0! * (1 - 0)!} * \dfrac{1}{t + 1} = WHAT?[/MATH]
And [MATH]n = 1 \implies (t + 1)^{-n} - 1 = (t + 1)^{-1} - 1 = \dfrac{1}{t + 1} - 1.[/MATH]
How would you prove those two are equal to each other?
Things equal to the same thing are equal to each other.
i know guys you are totally right...
this is what i did to prove that both sides are equal
Dr.Peterson
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Good.
Now, if you want to use my suggested approach rather than induction, try rewriting the summand in the general case, and observe that it contains powers of [MATH]\frac{-t}{t+1}[/MATH]. Can you relate it to the binomial theorem? (Have you learned that theorem???)
Or, if you think you need to use induction, show your attempt at the next step.
Now, if you want to use my suggested approach rather than induction, try rewriting the summand in the general case, and observe that it contains powers of [MATH]\frac{-t}{t+1}[/MATH]. Can you relate it to the binomial theorem? (Have you learned that theorem???)
Or, if you think you need to use induction, show your attempt at the next step.