Can someone help me with this
- Thread starter X049
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Steven G
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Is this a joke? You want a helper on a math help forum to do your work for you? I assure you that no helper here would think that would be helpful to you. Read the posting guidelines, post back following those guidelines and get the best help available.
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Look, I just want someone to point me in the right direction. Like in the first problem, there are 4 different values for x because x ≠ 1, so there's more than one value for the first sum. Therefore I think I'm doing something wrong and I don't know how to start.
D
Deleted member 4993
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a + a*x + a*x2 + a*x3 + a*x4 + .... +a*xn-1 + x*n = ?
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
Please share your work/thoughts about this problem
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@X049
What you say is correct for the second sum, so I wonder have they made a mistake with their indices in the second one.
They may mean [MATH]x^{2021}=1[/MATH] and mean to finish with [MATH]\frac{1}{1+x^{2020}}[/MATH], to mirror the first question.
In any case, there is one answer for all 5 roots in the first part (but you can find it trivially for x=1).
Looking at the first one, you should be able to complete the next one.
[MATH]\Sigma=\frac{1}{1+x}+\frac{1}{1+x^2}+\frac{1}{1+x^3}+\frac{1}{1+x^4}[/MATH]
Now [MATH]x^5=1[/MATH], so everywhere you see a 1, write [MATH]x^5[/MATH]
[MATH]\Sigma=\frac{x^5}{x^5+x}+\frac{x^5}{x^5+x^2}+\frac{x^5}{x^5+x^3}+\frac{x^5}{x^5+x^4}[/MATH]
[MATH]\Sigma=\frac{x^4}{1+x^4}+\frac{x^3}{1+x^3}+\frac{x^2}{1+x^2}+\frac{x}{1+x}[/MATH]
[MATH]\Sigma=1-\frac{1}{1+x^4}+1-\frac{1}{1+x^3}+1-\frac{1}{1+x^2}+1-\frac{1}{1+x}[/MATH]
[MATH]\Sigma=4-\Sigma[/MATH]
[MATH]\therefore \Sigma=2[/MATH]
What you say is correct for the second sum, so I wonder have they made a mistake with their indices in the second one.
They may mean [MATH]x^{2021}=1[/MATH] and mean to finish with [MATH]\frac{1}{1+x^{2020}}[/MATH], to mirror the first question.
In any case, there is one answer for all 5 roots in the first part (but you can find it trivially for x=1).
Looking at the first one, you should be able to complete the next one.
[MATH]\Sigma=\frac{1}{1+x}+\frac{1}{1+x^2}+\frac{1}{1+x^3}+\frac{1}{1+x^4}[/MATH]
Now [MATH]x^5=1[/MATH], so everywhere you see a 1, write [MATH]x^5[/MATH]
[MATH]\Sigma=\frac{x^5}{x^5+x}+\frac{x^5}{x^5+x^2}+\frac{x^5}{x^5+x^3}+\frac{x^5}{x^5+x^4}[/MATH]
[MATH]\Sigma=\frac{x^4}{1+x^4}+\frac{x^3}{1+x^3}+\frac{x^2}{1+x^2}+\frac{x}{1+x}[/MATH]
[MATH]\Sigma=1-\frac{1}{1+x^4}+1-\frac{1}{1+x^3}+1-\frac{1}{1+x^2}+1-\frac{1}{1+x}[/MATH]
[MATH]\Sigma=4-\Sigma[/MATH]
[MATH]\therefore \Sigma=2[/MATH]
Dr.Peterson
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Jomo was (rudely) asking you to tell us what help you need, by either showing your thoughts or asking a specific question. You've now more or less done that.
Lex has offered an idea (in fact, a full solution to the first part). Here's another suggestion:
First, the fact that x might be one of 4 different values doesn't necessarily imply that the sum has more than one value! In fact, the question hints that there might be something interesting to investigate about the sum. (When a problem looks wrong, maybe it's just amazing. Trust the author and give it a try.)
Playing with various ideas, I wondered if any pairs of terms might add up to a constant. Think about adding the first and last term, and then the two middle terms. What happens in each case?
@Dr.Peterson
Yes, the method I used shows that the first term is 1-(last term) etc...like your own method.
In your own method, to deal with an odd number of terms it's worth writing a second copy of the sum in reverse order below and adding first and last etc.. to get twice the sum - as in the method I used.
A curious, but interesting question.
Yes, the method I used shows that the first term is 1-(last term) etc...like your own method.
In your own method, to deal with an odd number of terms it's worth writing a second copy of the sum in reverse order below and adding first and last etc.. to get twice the sum - as in the method I used.
A curious, but interesting question.
Steven G
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How can you say that. You didn't even write anything. You just posted an image. From an image you want me to assume that you're asking for a hint. I was born at night, just not last night.
Steven G
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How do you know that just because you plug in different x-values that there will be more than one sum?
D
Deleted member 4993
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yeah i admit I wasn't very specific with my doubts sorry. And thanks for the other suggestion i will think about it.
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