­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2

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Thank you! Fixed the formatting in couple of places.

 


­Since \(1 + x + x^2 + ... + x^{n - 2} + x^{n - 1} =\frac{x^n-1}{x-1}\), then:

\(1 + 7 + 7^2 + ... +7^8 = 6,725,601\)

\(\frac{7^9-1}{7-1} = 6,725,601\)

\(\frac{7^9-1}{6} = 6,725,601\)

\(7^9-1= 6(6,725,601)\)

\(7^9= 6(6,725,601) + 1\)

Answer: A.

Hope it's clear.
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Hi Bunuel 

Many thanks for the solution and for directing me to this post, when I posted my own question before. Just as a potential tip: Maybe the topic name for this question can be renamed to "If x is NOT equal to 1...." otherwise this post is very hard to find. At least I did not manage to find it even after typing in the whole question.

Now regarding the question itself: Is the key to solving this question to recognize that the expression "\(1 + 7 + 7^2 + ... +7^8 = 6,725,601\)" is based on the fact that \(x^{n} \) in this case = \(7^{9}\)? ­and this in turn can be deduced from the last term in the expression \(7^{8}\) which stands for \(x^{n - 1}\) hence ­\(x^{n}\) itself must equal \(7^{9}\)­­? Then from here we can solve for \(7^{9}\) as you described in your solution above. I apologize if I might ask an obvious question but to me the wording was relatively confusing and I just want to make sure that I properly understand this exercise. 

Many thanks!­­­

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Yes, your understanding is correct. The expression implies that x^n equals 7^9, deduced from the last term in the sequence. From there, we can solve for 7^9 as demonstrated in the solution provided.