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

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The sequence S_n is such that, for every n where n > 1, S_n=(S_{n-1}-1 [#permalink]
Bunuel wrote:
The sequence $$S_n$$ is such that, for every n where n > 1, $$S_n=(S_{n-1}-1)^2$$. If $$S_5 = 100$$, what is $$S_3$$?

(A) 10
(B) 11
(C) √11
(D) √11 + 1
(E) √101

Kudos for a correct solution.

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

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:

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

$$\sqrt{11} = S_3 – 1$$

$$S_3 = \sqrt{11} + 1$$

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Re: The sequence S_n is such that, for every n where n > 1, S_n=(S_{n-1}-1 [#permalink]
Bunuel wrote:
The sequence $$S_n$$ is such that, for every n where n > 1, $$S_n=(S_{n-1}-1)^2$$. If $$S_5 = 100$$, what is $$S_3$$?

(A) 10
(B) 11
(C) √11
(D) √11 + 1
(E) √101

Kudos for a correct solution.

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

$$S3 = \sqrt{11} + 1$$
Re: The sequence S_n is such that, for every n where n > 1, S_n=(S_{n-1}-1 [#permalink]
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