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


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S1 = 1
S2 = - 1/2
S3 = -2
S4= 1
S5 = - 1/2
S6 = -2

and so on, it repeats in threes

therefore sum of first 61 terms = 20 * (1-1/2-2) + 1 = - 29

Ans (C)

MANHATTAN GMAT OFFICIAL SOLUTION:

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:

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

The correct answer is C.
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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
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\(S_n=\frac{-1}{S_{n-1}+1}\)
\(S_1\)= 1
\(S_2\)= -1/(1+1) = -1/2
\(S_3\)= -1/((-1/2)+1) = -2
\(S_4\)= -1/(-2+1) = 1

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

Answer C
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