G1.4 Given a sequence of real numbers $a_{1}, a_{2}, a_{3}, \ldots$ that satisfy
1) $a_{1}=\frac{1}{2}$, and
2) $a_{1}+a_{2}+\ldots+a_{k}=k^{2} a_{k}$, for $k \geq 2$.
Determine the value of $a_{100}$.
$$
\begin{array}{l}
\frac{1}{2}+a_{2}=2^{2} a_{2} \Rightarrow a_{2}=\frac{1}{2 \times 3}=\frac{1}{6} \\
\frac{1}{2}+\frac{1}{6}+a_{3}=3^{2} a_{3} \Rightarrow a_{3}=\frac{1}{3 \times 4}=\frac{1}{12}
\end{array}
$$

Claim: $a_{n}=\frac{1}{n \times(n+1)}$ for $n \geq 1$
Pf: By M.I., $n=1,2,3$, proved already.
Suppose $a_{k}=\frac{1}{k \times(k+1)}$ for some $k \geq 1$
$$
\frac{1}{2}+\frac{1}{6}+\ldots+\frac{1}{k \times(k+1)}+a_{k+1}=(k+1)^{2} a_{k+1}
$$
$\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{k}-\frac{1}{k+1}\right)=\left(k^{2}+2 k\right) a_{k+1}$
$a_{k+1}=\frac{1}{k(k+2)} \cdot\left(1-\frac{1}{k+1}\right)=\frac{1}{(k+1)(k+2)} \quad \therefore$ The statement is true for all $n \geq 1$
$$
a_{100}=\frac{1}{100 \times 101}=\frac{1}{10100}
$$