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# Is 1+x+x^2+x^3+....+x^10 positive?

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Senior Manager
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02 Jan 2011, 02:34
3
9
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Difficulty:

55% (hard)

Question Stats:

63% (01:48) correct 37% (01:41) wrong based on 287 sessions

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Is $$1+x+x^2+x^3+....+x^{10}$$positive?

(1) $$x<-1$$
(2) $$x^2>2$$

Source: GMAT Club Hardest problems.

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02 Jan 2011, 03:14
3
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gmatpapa wrote:
Is $$1+x+x^2+x^3+....+x^{10}$$positive?

1. $$x<-1$$
2. $$x^2>2$$

Source: GMAT Club Hardest problems.

Is $$1+(x+x^2)+(x^3+x^4)+...+(x^9+x^{10})>0$$?

(1) $$x<-1$$: $$x+x^2>0$$ (x<-1 meas that x^2>|x|), $$x^3+x^4>0$$, ..., $$x^9+x^{10}>0$$, so the sum is also more than zero. Sufficient.

(2) $$x^2>2$$: even if x itself is negative then still as above: $$x+x^2>0$$, $$x^3+x^4>0$$, ..., $$x^9+x^{10}>0$$, so the sum is also more than zero. Sufficient.

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03 Jan 2011, 11:34
Its clear now! Thanks!!
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04 Jan 2011, 06:56
Hi!
This is a geometric progression. Could you explain this using the sum formula?

Thanks!
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04 Jan 2011, 07:36
2
amankalra wrote:
Hi!
This is a geometric progression. Could you explain this using the sum formula?

Thanks!

Sum of the first $$n$$ terms of geometric progression is given by: $$sum=\frac{b*(r^{n}-1)}{r-1}$$, where $$b$$ is the first term, $$n$$ # of terms and $$r$$ is a common ratio $$\neq{1}$$.

So, in our case $$b=x$$ and $$r=x$$ --> $$1+(x+x^2+x^3+x^4+...+x^9+x^{10})=1+sum_{10}=1+\frac{x*(x^{10}-1)}{x-1}$$.

(1) $$x<-1$$ --> $$1+\frac{x*(x^{10}-1)}{x-1}=1+\frac{negative*positive}{negative}=1+positive=positive$$. Sufficient.

(2) $$x^2>2$$ --> $$x<-\sqrt{2}$$ or $$x>\sqrt{2}$$ --> so, either $$1+\frac{x*(x^{10}-1)}{x-1}=1+\frac{negative*positive}{negative}=1+positive=positive$$ or $$1+\frac{x*(x^{10}-1)}{x-1}=1+\frac{positive*positive}{positive}=1+positive=positive$$. Sufficient.

Hope it's clear.
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04 Jan 2011, 07:54
Thanks!
I was actually considering a total of 11 terms, and b=1.
Is that right?
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04 Jan 2011, 08:04
1
amankalra wrote:
Thanks!
I was actually considering a total of 11 terms, and b=1.
Is that right?

You can do that. If you take 1 as the first term then the formula will be $$sum=\frac{b*(r^{n}-1)}{r-1}=\frac{1*(x^{11}-1)}{x-1}=\frac{x^{11}-1}{x-1}$$.

(1) $$x<-1$$ --> $$\frac{x^{11}-1}{x-1}=\frac{negative}{negative}=positive$$. Sufficient.

(2) $$x^2>2$$ --> $$x<-\sqrt{2}\approx{-1.4}$$ or $$x>\sqrt{2}\approx{1.4}$$ --> so, either $$\frac{x^{11}-1}{x-1}=\frac{negative}{negative}=positive$$ or $$\frac{x^{11}-1}{x-1}=\frac{positive}{positive}=positive$$. Sufficient.

Hope it's clear.
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04 Jan 2011, 08:08
Okay.
Thanks a ton!
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04 Jan 2011, 21:14
I went for the 11 second solution and picked A...

If I took time on it and actualyl solved it - it is clear that the answer is D.

I got tripped up be thinking to myself "x could have two values -> not sufficient"

solution would have been let x = 2 or -2. -2 or 2^10 = 1024. 11 terms in series. Avg value of terms is 1025 => sum is 1025 / 2 = positive -> Sufficient

WHY OH WHY DO I RUSH?!?!?!
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25 Jan 2013, 06:05
Is 1 + x + x^2 + … + x^10 positive?

1) x < -1
2) x^2 > 2
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25 Jan 2013, 06:34
alexpavlos wrote:
Is 1 + x + x^2 + … + x^10 positive?

1) x < -1
2) x^2 > 2

Merging similar topics. Please refer to the solutions above.
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26 Jan 2013, 05:40
1
This is another one of those weird problems, where neither (1) nor (2) are necessary.

The expression 1 + x + xˆ2+...+ x^10 is ALWAYS positive, for any real value of x. PERIOD.

The best way to prove this, is transforming the expression in the GP sum (xˆ11 - 1)/(x-1) already mentioned.
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27 Apr 2014, 10:59
Bunuel wrote:
gmatpapa wrote:
Is $$1+x+x^2+x^3+....+x^{10}$$positive?

1. $$x<-1$$
2. $$x^2>2$$

Source: GMAT Club Hardest problems.

Is $$1+(x+x^2)+(x^3+x^4)+...+(x^9+x^{10})>0$$?

(1) $$x<-1$$: $$x+x^2>0$$ (x<-1 meas that x^2>|x|), $$x^3+x^4>0$$, ..., $$x^9+x^{10}>0$$, so the sum is also more than zero. Sufficient.

(2) $$x^2>2$$: even if x itself is negative then still as above: $$x+x^2>0$$, $$x^3+x^4>0$$, ..., $$x^9+x^{10}>0$$, so the sum is also more than zero. Sufficient.

Hi Bunnel,

According to st1 (x<-1)

so can write this as

1+ (-2) + (-2)^2+ (-2)^3+ ---(-2)^10

can I use above as geometric progression. ( dont include 1 in seried we will add it lastly)

If I will use this in gp then I will get result as <0 bcz first term is -ve

Thanks
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27 Apr 2014, 13:24
I agree with @caioguima, this is not a very good problem. GMAT won't give you a problem where the statements are completely redundant, and the answer to the question is already a definite Yes. At least, I haven't seen an official GMAT question that follows this format, please correct me if anyone has seen such an example.

To expand on @caioguima, the expression 1+x+x^2+x^3+....+x^10 is positive for all values of x greater than or equal to zero. If we rewrite this expression using the geometric series format, it becomes (x^11-1)/(x-1)[skipping those details here], and if we now consider the case of x<0, then both numerator and denominator are negative, making the expression positive for all values of x<0. Therefore, 1+x+x^2+x^3+....+x^10 is positive for all values of x. And the statements become redundant at this stage, which I have never seen on the GMAT.

Cheers,
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05 Jul 2019, 00:34
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Re: Is 1+x+x^2+x^3+....+x^10 positive?   [#permalink] 05 Jul 2019, 00:34
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