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

Question

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)

https://gmatclub.com/forum/download/file.php?id=75226

Bunuel · Accepted Answer

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.

Laracbo · Answer

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!

Bunuel · Answer

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.

kayarat600 · Answer

Is there a solution to this that doesn't require me.to memorize formulas

Krunaal · Answer

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.

kayarat600 · Answer

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

ExoticNeko · Answer

I don't understand why

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

Bunuel · Answer

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}

ExoticNeko · Answer

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 · Answer

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).

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

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

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