Loser66
  • Loser66
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.
Mathematics
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Loser66
  • Loser66
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.
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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

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ganeshie8
  • ganeshie8
looks there are several typoes with - and + in the main question could you pls double check
Loser66
  • Loser66
\[1+x+x^2 +\cdots+x^n =\sum_{j=0}^n x^j =\dfrac{x^{n+1}-1}{x-1}\] \[1+x+x^2 +\cdots +x^n+x^{n-1} = (\sum_{j=0}^n x^j) + x^{n-1}\\=\dfrac{x^{n+1}-1}{x-1}+x^{n-1}\] Simplify it, I get \(\dfrac{(x^{n+1}-1)+(x^{n-1}(x-1))}{x-1}\) open parentheses of the numerator: \[\dfrac{x^{n+1} -1+x^n -x^{n-1}}{x-1}\]
Loser66
  • Loser66
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?
dan815
  • dan815
what do u mean
dan815
  • dan815
okay here is the geometric series formula
dan815
  • dan815
|dw:1441205497301:dw|
dan815
  • dan815
|dw:1441205576147:dw|
Loser66
  • Loser66
yes
dan815
  • dan815
|dw:1441205592712:dw|
Loser66
  • Loser66
I got that part, dan
dan815
  • dan815
http://prntscr.com/8brzza
Loser66
  • Loser66
but that is the sum of 1 + x+ x^2 +....+x^n how about 1+x +x^2+.... +x^(n-1)?
dan815
  • dan815
1+x +x^2+.... +x^(n-1) = (X^n-1)/(x-1)
Loser66
  • Loser66
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!!
dan815
  • dan815
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
Loser66
  • Loser66
yes, sir
dan815
  • dan815
no problem mam
Loser66
  • Loser66
what do you mean by mam?
dan815
  • dan815
hahahhaha
Loser66
  • Loser66
i do "hate" you man!!

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