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# For all positive integers n, the nth term in sequence S_n is defined

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Math Expert
Joined: 02 Sep 2009
Posts: 58465
For all positive integers n, the nth term in sequence S_n is defined  [#permalink]

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10 Sep 2015, 00:30
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72% (01:44) correct 28% (01:43) wrong based on 189 sessions

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For all positive integers n, the nth term in sequence $$S_n$$ is defined as follows:

$$S_n = (n!)^{-1}$$

The sum of the first six terms of $$S_n$$ is

(A) between 0 and ½
(B) between ½ and 1
(C) between 1 and 1½
(D) between 1½ and 2
(E) greater than 2

Kudos for a correct solution.

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For all positive integers n, the nth term in sequence S_n is defined  [#permalink]

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10 Sep 2015, 01:30
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The sequence for the first six terms is: 1, 1/2, 1/6, 1/24, 1/120, 1/720

You can immediately rule out answer A, B and C as 1 + 1/2 alone will yield a result greater than 1,5.

Now the proceeding terms will get smaller and smaller and will not be 0,5 if summed, therefore the total sum has to be somewhere between 1,5 and 2. D
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Re: For all positive integers n, the nth term in sequence S_n is defined  [#permalink]

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10 Sep 2015, 04:58
Solution : Multiply and divide the sum by 6!
1/720(720+360+120+30+6+1)==> clearly b/n 1.5-2

Option D
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Re: For all positive integers n, the nth term in sequence S_n is defined  [#permalink]

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10 Sep 2015, 06:59
Bunuel wrote:
For all positive integers n, the nth term in sequence $$S_n$$ is defined as follows:

$$S_n = (n!)^{-1}$$

The sum of the first six terms of $$S_n$$ is

(A) between 0 and ½
(B) between ½ and 1
(C) between 1 and 1½
(D) between 1½ and 2
(E) greater than 2

Kudos for a correct solution.

The sum of sequencee of first 6 terms of $$S_n$$ will be

1+ $$\frac{1}{2!}$$ + $$\frac{1}{3!}$$ + $$\frac{1}{4!}$$ + $$\frac{1}{5!}$$ + $$\frac{1}{6!}$$

We do not actually need to solve for the sum.

The first two terms are 1 and 1/2. So the sum should be definitely greater than 1½ .
For the sum to be greater than or equal to 2, the sum of next fractions should be at least 1/2.
Now the succeeding fractions are less than 1/2, hence irrespective of the fraction, the sum will be less than 1/2.
So, sum of the sequence will be greater than 1½ but not more than 2.

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For all positive integers n, the nth term in sequence S_n is defined  [#permalink]

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10 Sep 2015, 10:31
1
Bunuel wrote:
For all positive integers n, the nth term in sequence $$S_n$$ is defined as follows:

$$S_n = (n!)^{-1}$$

The sum of the first six terms of $$S_n$$ is

(A) between 0 and ½
(B) between ½ and 1
(C) between 1 and 1½
(D) between 1½ and 2
(E) greater than 2

Kudos for a correct solution.

from the sequence equation we can see first 6 terms to be $$1+ \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} +\frac{1}{720}$$
So from 1st two numbers we have sum 1.5 and adding other four will be give more than 1.5 so A, B , C gone.
From D and E we can easily select D as from last four terms we have $$\frac{1}{6} (1+ \frac{1}{4} + \frac{1}{20} +\frac{1}{120})$$ which is less than .5. Hence D
Math Expert
Joined: 02 Sep 2009
Posts: 58465
Re: For all positive integers n, the nth term in sequence S_n is defined  [#permalink]

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13 Sep 2015, 09:36
1
Bunuel wrote:
For all positive integers n, the nth term in sequence $$S_n$$ is defined as follows:

$$S_n = (n!)^{-1}$$

The sum of the first six terms of $$S_n$$ is

(A) between 0 and ½
(B) between ½ and 1
(C) between 1 and 1½
(D) between 1½ and 2
(E) greater than 2

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

First, list out the first few terms of $$S_n$$ to get a feeling for the sequence.

$$S_1 = (1!)^{-1} = \frac{1}{1!} = \frac{1}{1} = 1$$
$$S_2 = (2!)^{-1} = \frac{1}{2!} = \frac{1}{2}$$

The sum of just the first two terms, by the way, is 1½. Glancing ahead at the answers, we should already be able to rule out A, B, and C at this point. The terms never go negative, so the only question is whether the sum of the first six terms takes us above 2 or not.

The sum of the first six terms can be written this way:

$$Sum = 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6!$$
$$= 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720$$

Given how quickly the fractions shrink, we might guess at this point that the sum does not go above 2. However, we can make sure of this guess. Leaving out the first two terms (the 1 and the ½), we can just figure out the sum of the last four terms:

$$1/6 + 1/24 + 1/120 + 1/720$$
$$= 1/6(1 + 1/4 + 1/20 + 1/120)$$
$$= 1/6(120/120 + 30/120 + 6/120 + 1/120)$$
$$= 1/6(157/120)$$
$$= 157/720$$,

which is definitely less than 1/2. (If it were above 1/2, then the sum of the first six terms would be greater than 2.)

Thus, the sum of the first six terms is between 1½ and 2.

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Re: For all positive integers n, the nth term in sequence S_n is defined  [#permalink]

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22 Feb 2019, 16:33
Bunuel wrote:
For all positive integers n, the nth term in sequence $$S_n$$ is defined as follows:

$$S_n = (n!)^{-1}$$

The sum of the first six terms of $$S_n$$ is

(A) between 0 and ½
(B) between ½ and 1
(C) between 1 and 1½
(D) between 1½ and 2
(E) greater than 2

Kudos for a correct solution.

S(1) = 1

S(2) = 1/2

S(3) = 1/6

S(4) = 1/24

S(5) = 1/120

S(6) = 1/720

Adding together the first 3 terms, we have 6/6 + 3/6 + 1/6 = 10/6 = 5/3 = 1 2/3.

Since 1/24 + 1/120 + 1/720 = 30/720 + 6/720 + 1/720 = 37/720 < 1/3, we see that the sum of the first six terms is between 1.5 and 2.

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Re: For all positive integers n, the nth term in sequence S_n is defined   [#permalink] 22 Feb 2019, 16:33
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