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Mathematics 93 Online
OpenStudy (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.

OpenStudy (loser66):

\(=\dfrac{(x-1)(x^{n+1})+(x^n-1)}{x-1}\\=\dfrac{x^{n+2}-x^{n+1}+x^n -1}{x-1}\)

OpenStudy (loser66):

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

OpenStudy (loser66):

@ganeshie8

ganeshie8 (ganeshie8):

looks there are several typoes with - and + in the main question could you pls double check

OpenStudy (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: \[\dfrac{x^{n+1} -1+x^n -x^{n-1}}{x-1}\]

OpenStudy (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?

OpenStudy (dan815):

what do u mean

OpenStudy (dan815):

okay here is the geometric series formula

OpenStudy (dan815):

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