The sequence S_n is such that, for every n where n > 1, S_n=(S_{n-1}-1

MANHATTAN GMAT OFFICIAL SOLUTION:

This problem is much easier if we can “decode” the sequence \(S_n=(S_{n-1}-1)^2\)

In other words, \(S_n\) is just any term in the sequence (for instance, \(S_3\) would be the 3rd term, \(S_10\) would be the 10th term, etc.)

\(S_{n-1}\) just means the term that comes right before \(S_n\).

Therefore, we can rephrase \(S_n=(S_{n-1}-1)^2\) as “To get any term, take the term before it, subtract one, then square.”

In this problem, however, we’ve been given \(S_5\) and we want \(S_3\). That is, we must go backwards.

To “go backwards” in a sequence, do the opposite of each step, in the opposite order. That is, if, to go from \(S_1\) to \(S_2\) you would:

1) subtract 1

2) square

…then to go backwards, you would:

1) square root

2) add 1

So, if \(S_5 = 100\), square root to get 10 and add 1 to get 11. Notice that we do not have to worry about the possibility of a negative square root, because every term is the square of some number, so no term can be negative.

If \(S_4 = 11\), square root to get √11 and add 1 to get √11 + 1. The answer is D.

The problem could also be solved a bit more algebraically as follows:

\(S_n=(S_{n-1}-1)^2\)

\(S_5 = 100\)

Therefore:

\(100 = (S_4-1)^2\)

\(10 = S_4 - 1\) (Again, we drop the possibility of a negative root, because \(S_4\) itself is a square.)

\(S_4 = 11\)

Now repeat the process:

\(11 = (S_3-1)^2\)

\(\sqrt{11} = S_3 – 1\)

\(S_3 = \sqrt{11} + 1\)

The correct answer is D.

\(S5 = (S4 - 1)^2 = ((S3-1)^2 - 1)^2 =100\)

\(S3 = \sqrt{11} + 1\)

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