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math_proof
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prove the same result, not by induction, but by directly manipulating the sum: let A be the sum, and show that xA = A + xn+1 1.(Use sigma notation in your proof).
 2 years ago
 2 years ago
math_proof Group Title
prove the same result, not by induction, but by directly manipulating the sum: let A be the sum, and show that xA = A + xn+1 1.(Use sigma notation in your proof).
 2 years ago
 2 years ago

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math_proof Group TitleBest ResponseYou've already chosen the best response.0
\[\sum_{i=0}^{n} x ^{i} = \frac{ 1x ^{n+1} }{ 1x }\]
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
First off, write (in sigma notation) what xA would be. What do you get?
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
i don't get it, whats xA
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
\[A=\sum_{i=0}^n x^i\]xA=?
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
im so confused
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
i hate does things
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Maybe it would help if you wrote the sum out. So you would get \[\sum_{i=0}^n x^i=x^0+x^1+x^2+x^3+...\]Then, \[xA=x\sum_{i=0}^n x^i=x(x^0+x^1+x^2+x^3+...)=x^1+x^2+x^3+x^4+...\]Can you write this back in sigma notation?
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
so thats gonna be the same as the above one
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Almost. There's one little difference. Your new sum should be \[\sum_{i=0}^n x^{i+1}\]All that's changed, is that each power of x has 1 added to it. Do you understand why this happens?
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
yeah i get it, but can't you change the index to start from 1? wouldn't it be the same?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
You could, you just need to be careful to also set the final index to n+1.That's why I prefer to start at 0. So \[\sum_{i=0}^n x^{i+1}=\sum_{i=1}^{n+1} x^i\]Now, all you need to do, is show that \(A+x^{n+1}1\) is the same thing as that new sum.
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
im having trouble understanding what the questions asks as to prove. whats A is the sum, then \[x \sum_{0}x ^{i}\]
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
so its basically xA=A+xn+1, but what about 1?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Notice that your new sum does not have the \(x^0=1\) term. So you need to subtract off the 1.
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
I don't get how its gonna equal \[\frac{ 1x ^{n+1} }{ 1x }\], how does it prove that? if I basically get \[\sum_{i=0}^{n} x ^{n+1}1\]
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Sorry it took so long for me to get back to you. OS was acting up for me quite seriously. But you want to show that \[\sum_{i=0}^n x^{i+1}=\left(\sum_{i=0}^n x^i\right)+x^{n+1}1\]There shouldn't be any division involved.
 2 years ago
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