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The recursive definition of Sn doesn't yield any secrets upon first glance. So let's write out the easy cases in the sequence, starting at n = 1 and going up until we notice a pattern:

Attachment:

2015-05-11_1808.png [ 36.58 KiB | Viewed 2001 times ]

etc. The terms of the sequence are 1, –1 /2 , –2 , 1, –1 /2, –2 … Three terms repeat in this cyclical pattern forever; every third term is the same. Now, the sum of each group of three consecutive terms is 1 + (–1/2) + (–2) = –3/2. There are 20 groups in the first 61 terms, with one term left over. So, the sum of the first 61 terms is:

(Number of groups)(Sum of one group) + (Leftover term) = (20)(–3/2) + 1 = –29

\(S_n\)= \(\frac{(-1)}{(S_{n-1}+1)}\) for all integer values of n greater than 1. If \(S_1\) = 1, what is the sum of the first 61 terms in the sequence?

whenever you see such Qs asking for sum of a large number of terms, more often than not , there will be a repetion in terms.. so lets see.. 1)\(S_1\) = 1 2)\(S_2\)= \(\frac{(-1)}{(S_1+1)}\)= \(\frac{-1}{2}\) 3) \(S_3\)= \(\frac{(-1)}{(S_2+1)}\)=-2 4)\(S_4\)= \(\frac{(-1)}{(S_3+1)}\)=1.. so we see after every three terms the numbers repeat.. so the sequence is-- 1, -1/2, -2, 1, -1/2.... sum of first three terms= 1-1/2 -2= -3/2.. so sum of first 60 terms= -3/2 * 60/3= -30.. 61 st term will be same as 1st so 1.. sum of all 61 terms = -30 + 1=-29.. C _________________

Re: S_n=-1/S_{n-1}+1 for all integer values of n greater than 1. If S1 = 1 [#permalink]

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09 Mar 2016, 03:59

The key to solving this question is to realise that that the terms are recursive and the cyclicity is 3 and we must add the 61st term which will be 1 hence => -1.5 x 20 +1 => -29
_________________

\(S_n=\frac{-1}{S_{n-1}+1}\) for all integer values of n greater than 1. If S1 = 1, what is the sum of the first 61 terms in the sequence?

(A) –48 (B) –31 (C) –29 (D) 1 (E) 30

Let’s determine the values of the first few terms of the sequence.

s(1) = 1

s(2) = -1/(1+1) = -½

s(3) = -1/(-½ + 1) = -1/(1/2) = -2

s(4) = -1/(-2 + 1) = -1/-1 = 1

We see that the terms repeat themselves in a cycle of 3 that has a pattern of 1, -1/2, -2, which sum to -1.5. We see that the sum of terms 1-3 is -1.5, and the sum of terms 4-6 is also -1.5, etc.

Thus, each set of 3 terms has a sum of -1.5. Since there are 20 sets of 3 from 1 to 60, inclusive, and the 61st term = 1, we can calculate the sum as follows:

20(-1.5) + 1 = -30 + 1 = -29.

Answer: C
_________________

Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

So, the sequence is 1, -1/2, -2, 1, -1/2...... The sequence repeats after every 3rd term i.e. \(S_1, S_4, S_7\)= 1 \(S_2, S_5, S_8\) = -1/2 \(S_3, S_6,S_9\)= -2

\(S_{61}\) = 1 Also \(S_1+S_2+S_3+......+S_{60}\) = 20*(1-1/2-2) = -30 So, \(S_1+S_2+S_3+......+S_{60}+S_{61}\) = -30 +1 = -29