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illegallyblonde
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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 27 Feb 2024, 09:46
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73% (01:52) correct 27% (02:08) wrong based on 1435 sessions

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­If x ≠ 1, then \(\frac{x^n-1}{x-1} = x^{n - 1} + x^{n - 2} + ... + x^2 + x + 1\). Given that \(1 + 7 + 7^2 + ... +7^8 = 6,725,601\), what does \(7^9\) equal?

A. 6(6,725,601) + 1
B. 6(6,725,601 + 1)
C. 8(6,725,601) + 1
D. (6,725,601 - 1)(6 + 1)
E. (6,725,601)(6 + 1)­­­­­­­­­

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Bunuel
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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 27 Feb 2024, 10:14
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helpmegmat
­If x ≠ 1, then \(\frac{x^n-1}{x-1} = x^{n - 1} + x^{n - 2} + ... +  x^2  +  x  +  1\). Given that \(1 + 7 + 7^2 + ... +7^8 = 6,725,601\), what does \(7^9\) equal?

A. 6(6,725,601) + 1
B. 6(6,725,601 + 1)
C. 8(6,725,601) + 1
D. (6,725,601 - 1)(6 + 1)
E. (6,725,601)(6 + 1)­­­­­­­­­

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­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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Laracbo
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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 03 Mar 2024, 16:37
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Bunuel
 
helpmegmat
­If x ≠ 1, then \(\frac{x^n-1}{x-1} = x^{n - 1} + x^{n - 2} + ... +  x^2  +  x  +  1\). Given that \(1 + 7 + 7^2 + ... +7^8 = 6,725,601\), what does \(7^9\) equal?

A. 6(6,725,601) + 1
B. 6(6,725,601 + 1)
C. 8(6,725,601) + 1
D. (6,725,601 - 1)(6 + 1)
E. (6,725,601)(6 + 1)­­­­­­­­­

Show Spoiler

­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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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 04 Mar 2024, 00:02
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Bunuel
 
helpmegmat
­If x ≠ 1, then \(\frac{x^n-1}{x-1} = x^{n - 1} + x^{n - 2} + ... +  x^2  +  x  +  1\). Given that \(1 + 7 + 7^2 + ... +7^8 = 6,725,601\), what does \(7^9\) equal?

A. 6(6,725,601) + 1
B. 6(6,725,601 + 1)
C. 8(6,725,601) + 1
D. (6,725,601 - 1)(6 + 1)
E. (6,725,601)(6 + 1)­­­­­­­­­

Show Spoiler

­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.
 ­­­
­

Hi Bunuel 

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.
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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 19 Jun 2025, 12:16
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Is there a solution to this that doesn't require me.to memorize formulas
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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 19 Jun 2025, 13:46
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kayarat600
Is there a solution to this that doesn't require me.to memorize formulas
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You don't need to memorize any formula to solve this question, simply substitute n - 1 = 8, and x = 7 looking at the given info. in question as done in this solution => https://gmatclub.com/forum/if-x-1-then- ... l#p3359219 and you will reach the answer.
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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 21 Jun 2025, 14:03
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kayarat600
Is there a solution to this that doesn't require me.to memorize formulas
You don't need to memorize any formula to solve this question, simply substitute n - 1 = 8, and x = 7 looking at the given info. in question as done in this solution => https://gmatclub.com/forum/if-x-1-then- ... l#p3359219 and you will reach the answer.
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Hi I'm guessing the latex for that equation doesn't load for me because I couldn't see the equation at all. Makes sense
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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 11 Dec 2025, 11:29
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I don't understand why

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



­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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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 11 Dec 2025, 11:38
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ExoticNeko
I don't understand why

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

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

Thus:

\(1 + 7 + 7^2 + ... +7^8 = \frac{7^9-1}{7-1} \)
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­If x 1, then (x^n - 1)/(x - 1) = x^{n - 1} + x^{n - 2} + ... +  x^2 11 Dec 2025, 11:53
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Can you rephrase this in another way? I still don't see the connection. between x^{n - 2} + x^{n - 1} and 7^8.
Bunuel

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

Thus:

\(1 + 7 + 7^2 + ... +7^8 = \frac{7^9-1}{7-1} \)
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ExoticNeko
Can you rephrase this in another way? I still don't see the connection. between x^{n - 2} + x^{n - 1} and 7^8.

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The formula says that the sum 1 + x + x^2 + ... + x^(n - 2) + x^(n - 1) is equal to (x^n - 1)/(x - 1).

We have the sum is 1 + 7 + 7^2 + ... + 7^8.

That means x = 7, and the highest power 8 must be n - 1, so n = 9. So the sum 1 + 7 + 7^2 + ... + 7^8 is exactly the pattern 1 + x + x^2 + ... + x^(n - 1) with x = 7 and n = 9, and therefore it equals (7^9 - 1)/(7 - 1).
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