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Re: Is 1+x+x^2+x^3+....+x^10 positive? [#permalink]
02 Jan 2011, 02:14
1
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Expert's post
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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.
Re: Is 1+x+x^2+x^3+....+x^10 positive? [#permalink]
04 Jan 2011, 06:36
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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 \(r=x\) --> \(1+(x+x^2+x^3+x^4+...+x^9+x^{10})=1+sum_{10}=1+\frac{x*(x^{10}-1)}{x-1}\).
(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.
Re: Is 1+x+x^2+x^3+....+x^10 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 \(sum=\frac{b*(r^{n}-1)}{r-1}=\frac{1*(x^{11}-1)}{x-1}=\frac{x^{11}-1}{x-1}\).
(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.
Re: Is 1+x+x^2+x^3+....+x^10 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
Re: Is 1+x+x^2+x^3+....+x^10 positive? [#permalink]
27 Apr 2014, 09:59
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
Re: Is 1+x+x^2+x^3+....+x^10 positive? [#permalink]
27 Apr 2014, 12:24
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.
Re: Is 1+x+x^2+x^3+....+x^10 positive? [#permalink]
15 May 2015, 19:23
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