Putting values of n, sequence turns out to be: 1, -1/2, -2, 1, -1/2...1

Sequence repeats it self every 3 terms so we can make 20 groups of the first three terms (for a total of 60 terms) and have the 61st term = 1 leftover

Sum of first three terms = -3/2

Sum of first 60 terms = 20*(-3/2) = -30

Adding the leftover 1: -30 + 1 = -29?

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)

Plug in the values into the formula:

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

As a result the sequence pattern is (1, - 1 / 2, -2).

61 terms divided by 3 equals in 20 1/3, so we need 20 times the sequence plus the first term of the pattern.

20 * (1 - 1/2 - 2) + 1 = - 29

Answer Choice 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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Hi debbiem,

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
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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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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
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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

Bunuel - what is a similar question to this? Where we can perhaps use (number of groups)(sum of one group) + uncounted terms?

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