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In the sequence a1,a2,a3...an, each term an is defined as an

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In the sequence a1,a2,a3...an, each term an is defined as an  [#permalink]

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New post 25 Aug 2014, 12:01
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In the sequence a1,a2,a3...an, each term an is defined as an=1/(−n)^n. The sum of the first ten terms of the sequence (a1,a2,a3...a10) must be __________.

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
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Re: In the sequence a1,a2,a3...an, each term an is defined as an  [#permalink]

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New post 25 Aug 2014, 12:05
Can anyone explain better than the solution below?

As with most sequence problems, the biggest step to take is to jot out the first few terms in search of a pattern or some other insight about the way the sequence unfolds. Here, that gives you:

a1=1/(−1)^1=−1

a2=1/(−2)^2=1/4

a3=1/(−3)^3=−1/27

a4=1/(−4)^4=1/256
What you should see here is that the odd terms are all negative and the evens are positive, and that the terms get so small so quickly as to be inconsequential. After just adding the first few terms you should see that you have a little less than −3/4 and that the remaining terms are going to be too small to swing that value very much in either direction, so the answer must be B.
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Re: In the sequence a1,a2,a3...an, each term an is defined as an  [#permalink]

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New post 27 Aug 2014, 11:21
I think the approach is fine. This question is more about understanding that the increasing squares in the denominator make the terms relatively inconsequential to the previous term - with respect to the answer choices.
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Re: In the sequence a1,a2,a3...an, each term an is defined as an  [#permalink]

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New post 27 Aug 2014, 17:21
ssagar wrote:
Can anyone explain better than the solution below?

As with most sequence problems, the biggest step to take is to jot out the first few terms in search of a pattern or some other insight about the way the sequence unfolds. Here, that gives you:

a1=1/(−1)^1=−1

a2=1/(−2)^2=1/4

a3=1/(−3)^3=−1/27

a4=1/(−4)^4=1/256
What you should see here is that the odd terms are all negative and the evens are positive, and that the terms get so small so quickly as to be inconsequential. After just adding the first few terms you should see that you have a little less than −3/4 and that the remaining terms are going to be too small to swing that value very much in either direction, so the answer must be B.


For those kinds of questions, I think the sum of the first 3 number will be sufficient to estimate the real sum.
The explanation for the method should be based on Taylor's theorem ( in my opinion)

I think for kinds of questions,
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Re: In the sequence a1,a2,a3...an, each term an is defined as an  [#permalink]

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New post 09 Mar 2016, 02:01
Somebody could give me the clear explanation?

I do not understand much about some ideas above.

Thank you.
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In the sequence a1,a2,a3...an, each term an is defined as an  [#permalink]

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New post 09 Mar 2016, 02:44
oanhnguyen1116 wrote:
Somebody could give me the clear explanation?

I do not understand much about some ideas above.

Thank you.


Hi,
Since there has been some confusion till now on this Q, let me give you a method
one approach that can be useful is--

Quote:
In the sequence a1,a2,a3...an, each term an is defined as an=1/(−n)^n. The sum of the first ten terms of the sequence (a1,a2,a3...a10) must be __________.

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


Solution--

\(a_n=\frac{1}{{-n}^n}\)



so
1) \(a_1=\frac{1}{{-1}^1}\)
2) \(a_2=\frac{1}{{-2}^2}\)..
3) \(a_3=\frac{1}{{-3}^3}\)...
4) \(a_4=\frac{1}{{-4}^4}\)..

INFERENCE:-


1)Starting from -1, alternate terms are -ive and positive..
2) Numeric value keeps decreasing as we go up..

so two scenarios--


1) (first term + second term) + (third term+fourth term) +... (ninth+tenth)..
=> \((-1+\frac{1}{4}) + (-\frac{1}{3^3} + \frac{1}{4^4}) ....+ (-\frac{1}{9^9} + \frac{1}{10^10})\)..
\(-\frac{3}{4} + (- ..) + (- ..)=- \frac{3}{4} - .. - ......\)
keeping the inference in mind, each pair will result in a negative qty
so the SUM will be lesser than \(-\frac{3}{4}\)..

2) lets take first term and then pair of 2nd-3rd, 4th-5th.. and so on
first term + (second term+third term)+(fourth term+fifth term)...(eighth+ninth)+tenth..
=>\(-1+ (\frac{1}{4}-\frac{1}{3^3} ) + ( \frac{1}{4^4}-..) .... (\frac{1}{8^8}-\frac{1}{9^9}) + \frac{1}{10^10}\)..
so leaving -1 and \(\frac{1}{10^10}\), all pair will result in a positive value..
so\(-1+\frac{1}{10^10} +\) some positive value + some positive value +...
so SUM has to be >-1..

Now we have two limits for the SUM..
\(SUM<-\frac{3}{4}\)and\(Sum>-1\)..
so it lies between -1 and -3/4..
choice B contains this RANGE..
so ans B. Between −1 and −1/2

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Re: In the sequence a1,a2,a3...an, each term an is defined as an  [#permalink]

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New post 09 Mar 2016, 03:52
chetan2u wrote:
oanhnguyen1116 wrote:
Somebody could give me the clear explanation?

I do not understand much about some ideas above.

Thank you.


Hi,
Since there has been some confusion till now on this Q, let me give you a method
one approach that can be useful is--

Quote:
In the sequence a1,a2,a3...an, each term an is defined as an=1/(−n)^n. The sum of the first ten terms of the sequence (a1,a2,a3...a10) must be __________.

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


Solution--

\(a_n=\frac{1}{{-n}^n}\)



so
1) \(a_1=\frac{1}{{-1}^1}\)
2) \(a_2=\frac{1}{{-2}^2}\)..
3) \(a_3=\frac{1}{{-3}^3}\)...
4) \(a_4=\frac{1}{{-4}^4}\)..

INFERENCE:-


1)Starting from -1, alternate terms are -ive and positive..
2) Numeric value keeps decreasing as we go up..

so two scenarios--


1) (first term + second term) + (third term+fourth term) +... (ninth+tenth)..
=> \((-1+\frac{1}{4}) + (-\frac{1}{3^3} + \frac{1}{4^4}) ....+ (-\frac{1}{9^9} + \frac{1}{10^10})\)..
\(-\frac{3}{4} + (- ..) + (- ..)=- \frac{3}{4} - .. - ......\)
keeping the inference in mind, each pair will result in a negative qty
so the SUM will be lesser than \(-\frac{3}{4}\)..

2) lets take first term and then pair of 2nd-3rd, 4th-5th.. and so on
first term + (second term+third term)+(fourth term+fifth term)...(eighth+ninth)+tenth..
=>\(-1+ (\frac{1}{4}-\frac{1}{3^3} ) + ( \frac{1}{4^4}-..) .... (\frac{1}{8^8}-\frac{1}{9^9}) + \frac{1}{10^10}\)..
so leaving -1 and \(\frac{1}{10^10}\), all pair will result in a positive value..
so\(-1+\frac{1}{10^10} +\) some positive value + some positive value +...
so SUM has to be >-1..

Now we have two limits for the SUM..
\(SUM<-\frac{3}{4}\)and\(Sum>-1\)..
so it lies between -1 and -3/4..
choice B contains this RANGE..
so ans B. Between −1 and −1/2



Fab work...
I too did it the same way and came to the conclusion that the sum must be between -1 and -3/4 . Though i must confess the options bamboozled me a bit...
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Re: In the sequence a1,a2,a3...an, each term an is defined as an  [#permalink]

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New post 12 Mar 2016, 17:30
Thank you chetan2u so much! It's clear for me now <3
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Re: In the sequence a1,a2,a3...an, each term an is defined as an  [#permalink]

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