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Is 1+x+x^2+x^3+....+x^10 positive?
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02 Jan 2011, 02:34
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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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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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02 Jan 2011, 03:14
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. Answer: D.
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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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03 Jan 2011, 11:34
Its clear now! Thanks!!
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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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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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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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04 Jan 2011, 07:36
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: \(sum=\frac{b*(r^{n}1)}{r1}\), where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\). So, in our case \(b=x\) and \(r=x\) > \(1+(x+x^2+x^3+x^4+...+x^9+x^{10})=1+sum_{10}=1+\frac{x*(x^{10}1)}{x1}\). (1) \(x<1\) > \(1+\frac{x*(x^{10}1)}{x1}=1+\frac{negative*positive}{negative}=1+positive=positive\). Sufficient. (2) \(x^2>2\) > \(x<\sqrt{2}\) or \(x>\sqrt{2}\) > so, either \(1+\frac{x*(x^{10}1)}{x1}=1+\frac{negative*positive}{negative}=1+positive=positive\) or \(1+\frac{x*(x^{10}1)}{x1}=1+\frac{positive*positive}{positive}=1+positive=positive\). Sufficient. Answer: D. Hope it's clear.
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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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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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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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04 Jan 2011, 08:04
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 \(sum=\frac{b*(r^{n}1)}{r1}=\frac{1*(x^{11}1)}{x1}=\frac{x^{11}1}{x1}\). (1) \(x<1\) > \(\frac{x^{11}1}{x1}=\frac{negative}{negative}=positive\). Sufficient. (2) \(x^2>2\) > \(x<\sqrt{2}\approx{1.4}\) or \(x>\sqrt{2}\approx{1.4}\) > so, either \(\frac{x^{11}1}{x1}=\frac{negative}{negative}=positive\) or \(\frac{x^{11}1}{x1}=\frac{positive}{positive}=positive\). Sufficient. Answer: D. Hope it's clear.
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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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04 Jan 2011, 08:08
Okay. Thanks a ton!



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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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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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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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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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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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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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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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26 Jan 2013, 05:40
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)/(x1) already mentioned.



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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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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. Answer: D. 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 Please clarify. Thanks



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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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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^111)/(x1)[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.
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Re: Is 1+x+x^2+x^3+....+x^10 positive?
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