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
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
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 farCan you show that the proposition is true for n = 1?
Please show us what you have tried and exactly where you are stuck.
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Please share your work/thoughts about this problem.
no theres nothing before sigma .Is that supposed to be a "t" before the sigma? If not, what is it?
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.
Did you evaluate the left side?? It isn't a sum; it's a single term.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
View attachment 21756
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.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.
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.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.
i know guys you are totally right...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.