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anonymous
 3 years ago
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).
anonymous
 3 years ago
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).

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{i=0}^{n} x ^{i} = \frac{ 1x ^{n+1} }{ 1x }\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i don't get it, whats xA

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

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Maybe 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so thats gonna be the same as the above one

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Almost. 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?

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

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0You 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so its basically xA=A+xn+1, but what about 1?

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

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

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry 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.
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