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math_proof

  • 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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  1. math_proof
    • 3 years ago
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    \[\sum_{i=0}^{n} x ^{i} = \frac{ 1-x ^{n+1} }{ 1-x }\]

  2. KingGeorge
    • 3 years ago
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    First off, write (in sigma notation) what xA would be. What do you get?

  3. math_proof
    • 3 years ago
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    i don't get it, whats xA

  4. KingGeorge
    • 3 years ago
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    \[A=\sum_{i=0}^n x^i\]xA=?

  5. math_proof
    • 3 years ago
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    im so confused

  6. math_proof
    • 3 years ago
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    i hate does things

  7. math_proof
    • 3 years ago
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    those

  8. KingGeorge
    • 3 years ago
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    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?

  9. math_proof
    • 3 years ago
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    so thats gonna be the same as the above one

  10. KingGeorge
    • 3 years ago
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    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?

  11. math_proof
    • 3 years ago
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    yeah i get it, but can't you change the index to start from 1? wouldn't it be the same?

  12. KingGeorge
    • 3 years ago
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    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.

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

  14. math_proof
    • 3 years ago
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    so its basically xA=A+xn+1, but what about -1?

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

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

  17. KingGeorge
    • 3 years ago
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    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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