Bunuel wrote:

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

Kudos for a correct solution.

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

\(S_1\)= 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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