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

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Is 1+x+x^2+x^3+....+x^10 positive? 1. x<-1 2. x^2>2 [#permalink]  02 Jan 2011, 01:34
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71% (01:59) correct 28% (01:17) wrong based on 77 sessions
Is 1+x+x^2+x^3+....+x^10positive?

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

Source: GMAT Club Hardest problems.
[Reveal] Spoiler: OA

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Re: Is the sum positive? [#permalink]  02 Jan 2011, 02:14
Expert's post
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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Re: Is the sum positive? [#permalink]  03 Jan 2011, 10:34
Its clear now! Thanks!!
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Re: Is the sum positive? [#permalink]  04 Jan 2011, 05:56
Hi!
This is a geometric progression. Could you explain this using the sum formula?

Thanks!
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Re: Is the sum positive? [#permalink]  04 Jan 2011, 06:36
Expert's post
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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Re: Is the sum positive? [#permalink]  04 Jan 2011, 06:54
Thanks!
I was actually considering a total of 11 terms, and b=1.
Is that right?
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Re: Is the sum positive? [#permalink]  04 Jan 2011, 07:04
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Expert's post
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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Re: Is the sum positive? [#permalink]  04 Jan 2011, 07:08
Okay.
Thanks a ton!
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Re: Is the sum positive? [#permalink]  04 Jan 2011, 20:14
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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is 1 + x + x^2 + … + x^10 positive [#permalink]  25 Jan 2013, 05:05
Is 1 + x + x^2 + … + x^10 positive?

1) x < -1
2) x^2 > 2
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Re: is 1 + x + x^2 + … + x^10 positive [#permalink]  25 Jan 2013, 05:34
Expert's post
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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Re: Is 1+x+x^2+x^3+....+x^10 positive? 1. x<-1 2. x^2>2 [#permalink]  26 Jan 2013, 04:40
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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Re: Is 1+x+x^2+x^3+....+x^10 positive? 1. x<-1 2. x^2>2   [#permalink] 26 Jan 2013, 04:40
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