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Sequence S is defined as Sn=Sn-1 + 1 +1/(Sn-1 + 1) for all n

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Sequence S is defined as Sn=Sn-1 + 1 +1/(Sn-1 + 1) for all n [#permalink] New post 19 Dec 2012, 06:18
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Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?

(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850
[Reveal] Spoiler: OA

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Last edited by daviesj on 19 Dec 2012, 23:09, edited 1 time in total.
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Re: Sequence S is defined as for all n > 121. [#permalink] New post 19 Dec 2012, 07:49
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daviesj wrote:
Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850

method to solve plz...


S_1 = 100

S_2 = \frac{101^2 + 1}{101} \approx 101 (Since 1 is negligible when compared to 101^2)

So, the series is almost an arithmetic progression with a=100, d=1,

We have got to find the sum of "n" terms where "n" is 16.

S_{16} = \frac{16}{2}*(2*100 + (16-1)*1)

= 8*215 = 1720

Answer is C.
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Re: Sequence S is defined as for all n > 121. [#permalink] New post 19 Dec 2012, 20:14
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daviesj wrote:
Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850

method to solve plz...


S1 has been given a big value i.e. 100 instead of the usual 0/1 etc. Why? Because 1/100 is negligible when added to 101

S_n = S_{n-1} + 1 + \frac{1}{S_{n-1}}

S_2 = S_{1} + 1 + \frac{1}{S_{1}} = 100 + 1 + \frac{1}{100} = 101 approx

S_3 = 101+ 1 + 1/101 = 102 approx
.
.
S_{16} = 115approx

S_1 + S_2 + ...S_{16} = 100 + 101 + 102 + ... 115 = 16*100 + 15*16/2 = 1720

The sum will be a little more than 1720.
Answer (c)
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Re: Sequence S is defined as for all n > 121. [#permalink] New post 19 Dec 2012, 21:07
VeritasPrepKarishma wrote:
daviesj wrote:
Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850

method to solve plz...


S1 has been given a big value i.e. 100 instead of the usual 0/1 etc. Why? Because 1/100 is negligible when added to 101

S_n = S_{n-1} + 1 + \frac{1}{S_{n-1}}

S_2 = S_{1} + 1 + \frac{1}{S_{1}} = 100 + 1 + \frac{1}{100} = 101 approx

S_3 = 101+ 1 + 1/101 = 102 approx
.
.
S_{16} = 115approx

S_1 + S_2 + ...S_{16} = 100 + 101 + 102 + ... 115 = 16*100 + 15*16/2 = 1720

The sum will be a little more than 1720.
Answer (c)


i would say the question is wrong or, atleast, not an exact GMAT type question...why to assume N as an integer...it is not specified in the question that n is an integer....n could be 1.2, 1.2,....etc for n>1 when n is not an integer.... i am thinking in the GMAT prospective...what is the source of this qtn?

Last edited by muralilawson on 19 Dec 2012, 21:53, edited 2 times in total.
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Re: Sequence S is defined as for all n > 121. [#permalink] New post 19 Dec 2012, 21:19
daviesj wrote:
Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850

method to solve plz...


S1 = 100

S2 = 100 + (1 + \frac{1}{101})
If you will notice 1 + \frac{1}{101} is approximately 1... S2 = 100 + 1 is approx. ~ 101

S3 = 101 + (1 + \frac{1}{101})
If you wil notice 1 + \frac{1}{101} is approximately 1... S3 = 101 + 1 approx. ~ 102

Sum = 16 * 100 + 1 + 2 + 3 + ... + 15 = 1600 + \frac{15(15+1)}{2} = 1600 + 120 = 1720

Answer: C
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Re: Sequence S is defined as for all n > 121. [#permalink] New post 19 Dec 2012, 21:52
muralilawson wrote:
VeritasPrepKarishma wrote:
daviesj wrote:
Sequence S is defined as Sn = (Sn-1 +1) + {1 / (Sn-1 +1)} for all n > 1. If S1 = 100, then which of the following must be true of Q, the sum of the first 16 terms of S?
(A) 1,600 ≤ Q ≤ 1,650
(B) 1,650 ≤ Q ≤ 1,700
(C) 1,700 ≤ Q ≤ 1,750
(D) 1,750 ≤ Q ≤ 1,800
(E) 1,800 ≤ Q ≤ 1,850

method to solve plz...


S1 has been given a big value i.e. 100 instead of the usual 0/1 etc. Why? Because 1/100 is negligible when added to 101

S_n = S_{n-1} + 1 + \frac{1}{S_{n-1}}

S_2 = S_{1} + 1 + \frac{1}{S_{1}} = 100 + 1 + \frac{1}{100} = 101 approx

S_3 = 101+ 1 + 1/101 = 102 approx
.
.
S_{16} = 115approx

S_1 + S_2 + ...S_{16} = 100 + 101 + 102 + ... 115 = 16*100 + 15*16/2 = 1720

The sum will be a little more than 1720.
Answer (c)


i would say the question is wrong or, atleast, not an exact GMAT question...why to assume N as an integer...it is not specified in the question that n is an integer....n could be 1.2, 1.2,....etc for n>1 when n is not an integer.... i am thinking in the GMAT prospective...


Since this is a PS question and not a DS question, we are free to make that assumption. "n" is only a subscript indicating the ordinal number of each term and hence can be taken to be integers.
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Re: Sequence S is defined as for all n > 121.   [#permalink] 19 Dec 2012, 21:52
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