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Loser66

  • one year ago

What is wrong with this? \(1 + x+x^2 +\cdots +x^n = \sum_{j =0}^n x^j\) Hence \(1+x +x^2 +\cdots +x^{n-1}\\= x^{n-1}+1+x +x^2+\cdots +x^n = x^{n-1}+\sum_{j=0}^n x^j = x^{n+1}+\dfrac{x^n -1}{x -1}\) continue on comment Please, explain me.

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  1. Loser66
    • one year ago
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    \(=\dfrac{(x-1)(x^{n+1})+(x^n-1)}{x-1}\\=\dfrac{x^{n+2}-x^{n+1}+x^n -1}{x-1}\)

  2. Loser66
    • one year ago
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    While when we go directly from \(1+x + x^2 +\cdots + x^{n-1}= \sum_{j =0}^n x^{j -1}=\dfrac{x^n -1}{x-1}\)

  3. Loser66
    • one year ago
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    @ganeshie8

  4. ganeshie8
    • one year ago
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    looks there are several typoes with - and + in the main question could you pls double check

  5. Loser66
    • one year ago
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    \[1+x+x^2 +\cdots+x^n =\sum_{j=0}^n x^j =\dfrac{x^{n+1}-1}{x-1}\] \[1+x+x^2 +\cdots +x^n+x^{n-1} = (\sum_{j=0}^n x^j) + x^{n-1}\\=\dfrac{x^{n+1}-1}{x-1}+x^{n-1}\] Simplify it, I get \(\dfrac{(x^{n+1}-1)+(x^{n-1}(x-1))}{x-1}\) open parentheses of the numerator: \[\dfrac{x^{n+1} -1+x^n -x^{n-1}}{x-1}\]

  6. Loser66
    • one year ago
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    But if I apply directly the formula \[\sum_{j=0}^n x^{j-1}= \dfrac{x^n -1}{x-1}\] it is different from the above. Why?

  7. dan815
    • one year ago
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    what do u mean

  8. dan815
    • one year ago
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    okay here is the geometric series formula

  9. dan815
    • one year ago
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    |dw:1441205497301:dw|

  10. dan815
    • one year ago
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    |dw:1441205576147:dw|

  11. Loser66
    • one year ago
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    yes

  12. dan815
    • one year ago
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    |dw:1441205592712:dw|

  13. Loser66
    • one year ago
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    I got that part, dan

  14. dan815
    • one year ago
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    http://prntscr.com/8brzza

  15. Loser66
    • one year ago
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    but that is the sum of 1 + x+ x^2 +....+x^n how about 1+x +x^2+.... +x^(n-1)?

  16. dan815
    • one year ago
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    1+x +x^2+.... +x^(n-1) = (X^n-1)/(x-1)

  17. Loser66
    • one year ago
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    hahaha.... I got you. I am dummy!! I interpret it as 1+x+x^2 +......+x^n then add one more term +x^(n-1) while it should be interpret as 1 + x+x^2 +....+x^(n-1) without the last term eeeeeeeeeeeh!! thank you very much, dan , the fake girl!!

  18. dan815
    • one year ago
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    yeah just remember if u are adding 1 more number to the 'sequence' 1+x+x^2 +......+x^n then + x^(n+1) not +x^(n-1) thats already in there

  19. Loser66
    • one year ago
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    yes, sir

  20. dan815
    • one year ago
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    no problem mam

  21. Loser66
    • one year ago
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    what do you mean by mam?

  22. dan815
    • one year ago
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    hahahhaha

  23. Loser66
    • one year ago
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    i do "hate" you man!!

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