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# For every integer m from 1 to 100, inclusive, the mth term

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Senior Manager
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For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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04 Mar 2012, 15:18
4
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Difficulty:

65% (hard)

Question Stats:

61% (02:26) correct 39% (02:45) wrong based on 256 sessions

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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.

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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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04 Mar 2012, 16:24
3
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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 $$sum=\frac{b}{1-r}$$, where $$b$$ is the first term.

So, in our case, since the # of terms is large enough, the sum will be very close to $$sum=\frac{-\frac{1}{4}}{1-\frac{1}{4}}=-\frac{1}{3}$$.

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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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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  [#permalink]

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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  [#permalink]

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19 Jun 2012, 05:00
1
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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Karishma
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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30 Oct 2012, 22:37
Why should we consider this as an infinite sequence?
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Posts: 51185
Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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30 Oct 2012, 22:43
Ousmane wrote:
Why should we consider this as an infinite sequence?

We are not saying that it's an infinite sequence. But if it were, then the sum would be -1/3 and since we have the sum of the large enough number of terms (100), then the actual sum would be very close to -1/3.
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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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  [#permalink]

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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:
$$\frac{a(1-r^n)}{1-r}$$
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  [#permalink]

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22 Nov 2017, 06:58
Bunuel Why don't we use the formula a(r^n-1)/r-1?
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Re: For every integer m from 1 to 100, inclusive, the mth term  [#permalink]

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22 Nov 2017, 07:32
Bunuel Why don't we use the formula a(r^n-1)/r-1?

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 [#permalink] 22 Nov 2017, 07:32
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