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# The sequence S is defined as follows for all n ≥ 1: The sum of the

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The sequence S is defined as follows for all n ≥ 1: The sum of the  [#permalink]

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17 Jun 2015, 06:00
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The sequence S is defined as follows for all n ≥ 1:

$$S_n=(-1)^n*\frac{1}{n(n+1)}$$

The sum of the first 10 terms of S is:

(A) Between –1 and –1/2
(B) Between –1/2 and 0
(C) Between 0 and 1/2
(D) Between 1/2 and 1
(E) Greater than 1

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Re: The sequence S is defined as follows for all n ≥ 1: The sum of the  [#permalink]

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17 Jun 2015, 06:33
1
Bunuel wrote:
The sequence S is defined as follows for all n ≥ 1:

$$S_n=(-1)^n*\frac{1}{n(n+1)}$$

The sum of the first 10 terms of S is:

(A) Between –1 and –1/2
(B) Between –1/2 and 0
(C) Between 0 and 1/2
(D) Between 1/2 and 1
(E) Greater than 1

Given: $$S_n=(-1)^n*\frac{1}{n(n+1)}$$

i.e. $$S_1=(-1)^1*\frac{1}{1(1+1)} = -1/2$$
i.e. $$S_2=(-1)^2*\frac{1}{2(2+1)} = 1/6$$
i.e. $$S_3=(-1)^3*\frac{1}{3(3+1)} = -1/12$$
i.e. $$S_4=(-1)^4*\frac{1}{4(4+1)} = 1/20$$
i.e. $$S_5=(-1)^5*\frac{1}{5(5+1)} = -1/30$$
... and so on

Sum of First 10 terms = (-1/2) + (1/6) + (-1/12) + (1/20) + (-1/30) + (1/42) .... and so on

Method-1

Substitute the approximate values of terms

Sum of First 10 terms = -0.5 + 0.16 - 0.083 + 0.05 - 0.033 + 0.023 - ...

Add every two terms together

Sum of First 10 terms = (-0.5 + 0.16) + (-0.083 + 0.05) + (-0.033 + 0.023) + ...
Sum of First 10 terms = (-0.34) + (-0.03) + (-0.01) + ... = approx. (-0.4) because the next terms will be very small

i.e. It falls between 0 and -0.5

Answer: Option
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Re: The sequence S is defined as follows for all n ≥ 1: The sum of the  [#permalink]

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17 Jun 2015, 07:01
Bunuel wrote:
The sequence S is defined as follows for all n ≥ 1:

$$S_n=(-1)^n*\frac{1}{n(n+1)}$$

The sum of the first 10 terms of S is:

(A) Between –1 and –1/2
(B) Between –1/2 and 0
(C) Between 0 and 1/2
(D) Between 1/2 and 1
(E) Greater than 1

Given: $$S_n=(-1)^n*\frac{1}{n(n+1)}$$

i.e. $$S_1=(-1)^1*\frac{1}{1(1+1)} = -1/2$$
i.e. $$S_2=(-1)^2*\frac{1}{2(2+1)} = 1/6$$
i.e. $$S_3=(-1)^3*\frac{1}{3(3+1)} = -1/12$$
i.e. $$S_4=(-1)^4*\frac{1}{4(4+1)} = 1/20$$
i.e. $$S_5=(-1)^5*\frac{1}{5(5+1)} = -1/30$$
... and so on

Sum of First 10 terms = (-1/2) + (1/6) + (-1/12) + (1/20) + (-1/30) + (1/42) .... and so on

Method-2

i.e. Sum of First 10 terms = (-1/2) + (1/12) + (1/60) + .... and so on

i.e. Sum of First 10 terms = (-1/2) + some very small positive values

i.e. Sum of First 10 terms = Greater that (-1/2) but less than Zero

i.e. It falls between 0 and -0.5

Answer: Option
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The sequence S is defined as follows for all n ≥ 1: The sum of the  [#permalink]

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17 Jun 2015, 07:23
1
1
Bunuel wrote:
The sequence S is defined as follows for all n ≥ 1:

$$S_n=(-1)^n*\frac{1}{n(n+1)}$$

The sum of the first 10 terms of S is:

(A) Between –1 and –1/2
(B) Between –1/2 and 0
(C) Between 0 and 1/2
(D) Between 1/2 and 1
(E) Greater than 1

Hi,
i would do this Q in two steps...
1) first step would be to eliminate all wrong answers..
as we can see the first term is negative ,next +ive and so on.. basically all odd numbers are -ive and even numbers are +ive..
Also the denominator is increasing in subsequent term, so the answer has to be a negative number that is <0..
only A and B are left..
2) second step to close on the answer between the two requires a bit of calculations...
$$\frac{1}{(n(n+1))}$$=$$\frac{1}{n}-\frac{1}{(n+1)}$$...
so first two terms give us=$$-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{3}=-\frac{1}{3}$$..
next two terms=$$-\frac{1}{3}+\frac{1}{4}+\frac{1}{4}-\frac{1}{5}=-\frac{1}{30}$$...
we can see the vaue gets significantly smaller with every next two terms so we can estimate that the answer will be slightly lesser than -1/3 but should be more than -1/2..
so ans B.. we can actually next two terms as -1/105.. and can find other two to get to exact answer but may not require that..
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Re: The sequence S is defined as follows for all n ≥ 1: The sum of the  [#permalink]

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17 Jun 2015, 08:08
The sequence S is defined as follows for all n ≥ 1:

$$S_n=(-1)^n*\frac{1}{n(n+1)}$$

The sum of the first 10 terms of S is:

(A) Between –1 and –1/2
(B) Between –1/2 and 0
(C) Between 0 and 1/2
(D) Between 1/2 and 1
(E) Greater than 1

Solution -

S1 = -1/2
S2 = 1/6
S3 = -1/12
S4 = 1/20 .....

Neglecting the remaining terms as fractions does not vary the sum more than 1%. [We do not have time to calculate all in GMAT test]

S=S1+S2+S3+S4...S10 > -1/3, but S<-1/4 .

-1/3 <S< -1/4. So answer falls in B.

Thanks,

Kudos Please.
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Re: The sequence S is defined as follows for all n ≥ 1: The sum of the  [#permalink]

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17 Jun 2015, 08:18
balamoon wrote:
The sequence S is defined as follows for all n ≥ 1:

$$S_n=(-1)^n*\frac{1}{n(n+1)}$$

The sum of the first 10 terms of S is:

(A) Between –1 and –1/2
(B) Between –1/2 and 0
(C) Between 0 and 1/2
(D) Between 1/2 and 1
(E) Greater than 1

Solution -

S1 = -1/2
S2 = 1/6
S3 = -1/12
S4 = 1/20 .....

Neglecting the remaining terms as fractions does not vary the sum more than 1%. [We do not have time to calculate all in GMAT test]

S=S1+S2+S3+S4...S10 > -1/3, but S<-1/4 .

-1/3 <S< -1/4
. So answer falls in B.

Thanks,

Kudos Please.

I think the highlighted part in your above mentioned explanation is INCORRECT.

The actual Sum of the series is -0.3821789 which does NOT satisfy the range -1/3 <S< -1/4
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Re: The sequence S is defined as follows for all n ≥ 1: The sum of the  [#permalink]

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17 Jun 2015, 11:45
1
lets check first 4 terms only ,
S1 = -1/2
S2 = 1/6
S3 = -1/12
S4 = 1/20

Add S1 and S2 , = -1/3
So sum of every term in series is negative so sum of entire first 10 term is negative ... this eliminates option C, D , E

Now S1+S2 = -1/3 = -0.333
after this terms are only going get smaller and smaller . The sum of next 8 terms can not shoot this value beyond -0.50 because then in that case sum (s3 +s4+ ..... +s10) should constiute more than 30% of (S1 to S10) , which looks imposible as values will get smaller and smaller only .

So total should lie between 0 and -1/2.
Answer B
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Re: The sequence S is defined as follows for all n ≥ 1: The sum of the  [#permalink]

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22 Jun 2015, 06:56
Bunuel wrote:
The sequence S is defined as follows for all n ≥ 1:

$$S_n=(-1)^n*\frac{1}{n(n+1)}$$

The sum of the first 10 terms of S is:

(A) Between –1 and –1/2
(B) Between –1/2 and 0
(C) Between 0 and 1/2
(D) Between 1/2 and 1
(E) Greater than 1

MANHATTAN GMAT OFFICIAL SOLUTION:

We should compute the first few elements of Sn. Because we need to know the sum of the first 10 elements, we should also track the cumulative sum:

By now (if not before!) the pattern should be fairly obvious. The sum of the first n terms of Sn converges somewhere in the range between –0.3667 and –0.4. Only (B) exhibits a range in which the sum of this series could converge.

The correct answer is B.

Attachment:

2015-06-22_1755.png [ 99.92 KiB | Viewed 2515 times ]

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Re: The sequence S is defined as follows for all n ≥ 1: The sum of the  [#permalink]

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30 Apr 2019, 15:38
Bunuel wrote:
The sequence S is defined as follows for all n ≥ 1:

$$S_n=(-1)^n*\frac{1}{n(n+1)}$$

The sum of the first 10 terms of S is:

(A) Between –1 and –1/2
(B) Between –1/2 and 0
(C) Between 0 and 1/2
(D) Between 1/2 and 1
(E) Greater than 1

sum = 1/2 last+first * number of term
i choose 1 and then put 10 and according to formula got -54/110 and my naswer is b
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Re: The sequence S is defined as follows for all n ≥ 1: The sum of the   [#permalink] 30 Apr 2019, 15:38
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