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- anonymous

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

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- anonymous

\[\sum_{i=0}^{n} x ^{i} = \frac{ 1-x ^{n+1} }{ 1-x }\]

- KingGeorge

First off, write (in sigma notation) what xA would be. What do you get?

- anonymous

i don't get it, whats xA

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- KingGeorge

\[A=\sum_{i=0}^n x^i\]xA=?

- anonymous

im so confused

- anonymous

i hate does things

- anonymous

those

- KingGeorge

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?

- anonymous

so thats gonna be the same as the above one

- KingGeorge

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?

- anonymous

yeah i get it, but can't you change the index to start from 1? wouldn't it be the same?

- KingGeorge

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.

- anonymous

im having trouble understanding what the questions asks as to prove. whats A is the sum, then \[x \sum_{0}x ^{i}\]

- anonymous

so its basically xA=A+xn+1, but what about -1?

- KingGeorge

Notice that your new sum does not have the \(x^0=1\) term. So you need to subtract off the 1.

- anonymous

I don't get how its gonna equal \[\frac{ 1-x ^{n+1} }{ 1-x }\], how does it prove that? if I basically get \[\sum_{i=0}^{n} x ^{n+1}-1\]

- KingGeorge

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.

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