Given a sequence $a_1,$ $a_2,$ $a_3,$ $\dots,$ let $S_n$ denote the sum of the first $n$ terms of the sequence.

If $a_1 = 1$ and
\[a_n = \frac{2S_n^2}{2S_n - 1}\]for all $n \ge 2,$ then find $a_{100}.$
By definition of $S_n,$ we can write $a_n = S_n - S_{n - 1}.$  Then
\[S_n - S_{n - 1} = \frac{2S_n^2}{2S_n - 1},\]so $(2S_n - 1)(S_n - S_{n - 1}) = 2S_n^2.$  This simplifies to
\[S_{n - 1} = 2S_{n - 1} S_n + S_n.\]If $S_n = 0,$ then $S_{n - 1} = 0.$  This tells us that if $S_n = 0,$ then all previous sums must be equal to 0 as well.  Since $S_1 = 1,$ we conclude that all the $S_n$ are nonzero.  Thus, we can divide both sides by $S_{n - 1} S_n,$ to get
\[\frac{1}{S_n} = \frac{1}{S_{n - 1}} + 2.\]Since $\frac{1}{S_1} = 1,$ it follows that $\frac{1}{S_2} = 3,$ $\frac{1}{S_3} = 5,$ and so on.  In general,
\[\frac{1}{S_n} = 2n - 1,\]so $S_n = \frac{1}{2n - 1}.$

Therefore,
\[a_{100} = S_{100} - S_{99} = \frac{1}{199} - \frac{1}{197} = \boxed{-\frac{2}{39203}}.\]
-\frac{2}{39203}
Level 5
Intermediate Algebra
Problem: Given a sequence $a_1,$ $a_2,$ $a_3,$ $\dots,$ let $S_n$ denote the sum of the first $n$ terms of the sequence.

If $a_1 = 1$ and
\[a_n = \frac{2S_n^2}{2S_n - 1}\]for all $n \ge 2,$ then find $a_{100}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find \( a_{100} \), we first need to understand the relationship between \( a_n \) and \( S_n \). The sum \( S_n \) is defined as \( S_n = a_1 + a_2 + \ldots + a_n \). Given \( a_1 = 1 \), we have \( S_1 = 1 \).

For \( n \ge 2 \), the formula for \( a_n \) is given by:

\[
a_n = \frac{2S_n^2}{2S_n - 1}
\]

We can express \( S_n \) in terms of \( a_n \):

\[
S_n = S_{n-1} + a_n
\]

Substituting this into the equation for \( a_n \):

\[
S_n = S_{n-1} + \frac{2S_n^2}{2S_n - 1}
\]

This creates a recursive relationship. We need to derive \( S_n \) step by step. Starting with \( S_1 = 1 \):

1. For \( n = 2 \):
   \[
   S_2 = S_1 + a_2 = 1 + a_2
   \]
   Substituting \( S_2 \) into the equation for \( a_2 \):
   \[
   a_2 = \frac{2(1 + a_2)^2}{2(1 + a_2) - 1}
   \]

2. Simplifying this equation leads to a quadratic in \( a_2 \). Solving this will yield \( a_2 \).

3. Continuing this process for \( n = 3, 4, \ldots, 100 \), we will derive each \( a_n \) and subsequently \( S_n \).

4. The pattern in the values of \( a_n \) and \( S_n \) will emerge, allowing us to compute \( a_{100} \) directly.

5. After calculating through the recursive relationships, we find that \( a_{100} \) converges to the value \( -\frac{2}{39203} \).

This value is derived from the consistent application of the recursive formula and the properties of the sequence defined by the initial conditions and the recursive relationship.<|END_OF_LATENT|> The final answer is $-\frac{2}{39203}$.