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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]

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New post Updated on: 20 Dec 2012, 00:09
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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

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Originally posted by daviesj on 19 Dec 2012, 07:18.
Last edited by daviesj on 20 Dec 2012, 00:09, edited 1 time in total.
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Re: Sequence S is defined as for all n > 121.  [#permalink]

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New post 19 Dec 2012, 08:49
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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]

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New post 19 Dec 2012, 21:14
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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
.
.
\(S_{16} = 115\)approx

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

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]

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New post Updated on: 19 Dec 2012, 22:53
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
.
.
\(S_{16} = 115\)approx

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

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?

Originally posted by muralilawson on 19 Dec 2012, 22:07.
Last edited by muralilawson on 19 Dec 2012, 22:53, edited 2 times in total.
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Re: Sequence S is defined as for all n > 121.  [#permalink]

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New post 19 Dec 2012, 22:19
2
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]

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New post 19 Dec 2012, 22:52
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
.
.
\(S_{16} = 115\)approx

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

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 Sn=Sn-1 + 1 +1/(Sn-1 + 1) for all n  [#permalink]

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New post 10 Nov 2016, 00:25
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


Answer: Option C

Please check the solution in attachment
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Re: Sequence S is defined as Sn=Sn-1 + 1 +1/(Sn-1 + 1) for all n  [#permalink]

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