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For every integer m from 1 to 100, inclusive, the mth term
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04 Mar 2012, 15:18
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For every integer m from 1 to 100, inclusive, the m th term of a certain sequence is given by (1)^m*2^(m). If N is the sum of the fi
rst 100 terms in the sequence, then N is (A) less than 1 (B) between 1 and 1/2 (C) between 1/2 and 0 (D) between 0 and 1/2 (E) greater than 1/2 Any idea how solve this question? I have got no clue how to.
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Re: For every integer m from 1 to 100, inclusive, the mth term
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04 Mar 2012, 16:24
enigma123 wrote: For every integer m from 1 to 100, inclusive, the mth term of a certain sequence is given by (1)^m*2^(m). If N is the sum of the fi
rst 100 terms in the sequence, then N is
(A) less than 1 (B) between 1 and 1/2 (C) between 1/2 and 0 (D) between 0 and 1/2 (E) greater than 1/2
Any idea how solve this question? I have got no clue how to. 1st term \((1)^1*2^{1}=\frac{1}{2}\); 2nd term \((1)^2*2^{2}=\frac{1}{4}\); 3rd term \((1)^3*2^{3}=\frac{1}{8}\); 4th term \((1)^4*2^{4}=\frac{1}{16}\); 5th term \((1)^5*2^{5}=\frac{1}{32}\); 6th term \((1)^6*2^{6}=\frac{1}{64}\); ... The sum of 1st and 2nd = \(\frac{1}{4}\); The sum of 3rd and 4th = \(\frac{1}{16}\); The sum of 5th and 6th = \(\frac{1}{64}\); ... So, we should basically sum 100/2=50 terms which form a geometric progression with the first term of \(\frac{1}{4}\) and common ratio of \(\frac{1}{4}\): \(\frac{1}{4}\), \(\frac{1}{16}\), \(\frac{1}{64}\), ... The sum of infinite geometric progression with common ratio \(r<1\), is \(sum=\frac{b}{1r}\), where \(b\) is the first term. So, in our case, since the # of terms is large enough, the sum will be very close to \(sum=\frac{\frac{1}{4}}{1\frac{1}{4}}=\frac{1}{3}\). Answer: C.
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Re: For every integer m from 1 to 100, inclusive, the mth term
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05 May 2012, 04:40
I don't get it. Isn't the first term 1/2 and the common ratio 1/2? And why do you need to sum up 1 & 2, 3 & 4, 5 & 6 term... etc.?
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Re: For every integer m from 1 to 100, inclusive, the mth term
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19 Jun 2012, 04:09
Stiv wrote: I don't get it. Isn't the first term 1/2 and the common ratio 1/2? And why do you need to sum up 1 & 2, 3 & 4, 5 & 6 term... etc.? Well both the solution would fetch you the same result of 1/3 if you consider it as an infinite series.. and that would solve the question.
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Re: For every integer m from 1 to 100, inclusive, the mth term
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19 Jun 2012, 05:00
enigma123 wrote: For every integer m from 1 to 100, inclusive, the mth term of a certain sequence is given by (1)^m*2^(m). If N is the sum of the fi
rst 100 terms in the sequence, then N is
(A) less than 1 (B) between 1 and 1/2 (C) between 1/2 and 0 (D) between 0 and 1/2 (E) greater than 1/2
Any idea how solve this question? I have got no clue how to. I have discussed this question in details using both the methods  GP perspective and non GP perspective. Check it out here: http://www.veritasprep.com/blog/2012/04 ... rspective/
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Re: For every integer m from 1 to 100, inclusive, the mth term
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30 Oct 2012, 22:37
Why should we consider this as an infinite sequence?



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Re: For every integer m from 1 to 100, inclusive, the mth term
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30 Oct 2012, 22:43



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Re: For every integer m from 1 to 100, inclusive, the mth term
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04 May 2013, 05:06
Ousmane wrote: Why should we consider this as an infinite sequence? Lets say we have only the first term m=1 , then the sum of the series (since it is just one term) , would be 1/2 Now , as explained by Bunuel , lets suppose there are infinite terms , in which case the sum is 1/3 . Hence sum of 1 to 100 terms SHOULD lie between these two range "1/2 and 1/3". Jyothi
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Re: For every integer m from 1 to 100, inclusive, the mth term
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05 May 2013, 08:45
A general term of the expression is: \((1)^m.2^{m}\) Hence, the 1st 100 terms can be written as: \(t_1=\frac{1}{2}\) \(t_2=\frac{1}{4}\) \(t_3=\frac{1}{8}\) . . . . . \(t_{100}=\frac{1}{2^{100}}\) Now, we know that the sum of the first "n" terms of a Geometric progression with a common ratio r<1 is: \(\frac{a(1r^n)}{1r}\) where a=1st term, r=common ratio and n=no. of terms. Now for this case the expression would be(after putting all the values): \(N=\frac{\frac{1}{2}[1(\frac{1}{2})^{100}]}{1(\frac{1}{2})}\) \(=\frac{\frac{1}{2}[1\frac{1}{2^{100}}]}{\frac{3}{2}}\) \(=\frac{1}{3}.[1\frac{1}{2^{100}}]\) Now, since \(2^{100}\) is a very large number compared to 1, Hence, we can cay that \(1(\frac{1}{2^{100}})\) is slightly less than 1 Hence, N is slightly less than \(\frac{1}{3}\).
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Re: For every integer m from 1 to 100, inclusive, the mth term
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22 Nov 2017, 06:58
Bunuel Why don't we use the formula a(r^n1)/r1?



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Re: For every integer m from 1 to 100, inclusive, the mth term
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22 Nov 2017, 07:32
radhikakhemka007 wrote: Bunuel Why don't we use the formula a(r^n1)/r1? We can use this formula but the result will be same i.e. 1/3
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Re: For every integer m from 1 to 100, inclusive, the mth term &nbs
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22 Nov 2017, 07:32






