3A. Given the sequence $a_{1}, a_{2}, \ldots, a_{n} \ldots$, defined by:

$$
a_{1}=2 \text { and } a_{n+1}=1+a_{1} a_{2} \ldots a_{n}
$$

for every natural number, $n \geq 2$. Calculate $S_{2009}+P_{2009}$, where

$$
S_{k}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}} \text { and } P_{k}=\frac{1}{a_{1} a_{2} \ldots a_{n}}
$$

for every $k \geq 1$.
Solution. Let $t_{n}=S_{n}+P_{n}$ then

$$
t_{1}=S_{1}+P_{1}=\frac{1}{2}+\frac{1}{2}=1, t_{2}=S_{2}+P_{2}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1
$$

We will show by mathematical induction that $t_{n}=1$, for every $n \in \mathbb{N}$. Note that

$$
S_{n+1}=S_{n}+\frac{1}{a_{n+1}} \text { and } P_{n+1}=\frac{P_{n}}{a_{n+1}}
$$

For $n=1$, the statement clearly holds.

Assume that for $n=k$ the statement holds, i.e., $t_{k}=1$.

For $n=k+1$ we have

$$
\begin{aligned}
t_{k+1} & =S_{k+1}+P_{k+1}=S_{k}+\frac{1}{a_{k+1}}+\frac{P_{k}}{a_{k+1}} \\
& =S_{k}+P_{k}+\frac{1}{a_{k+1}}+\frac{P_{k}}{a_{k+1}}-P_{k}
\end{aligned}
$$

$$
\begin{aligned}
& =1+\frac{1}{a_{k+1}}+\frac{P_{k}}{a_{k+1}}\left(1-a_{k+1}\right) \\
& =1+\frac{1}{a_{k+1}}+\frac{P_{k}}{a_{k+1}}\left(-a_{1} a_{2} \ldots a_{k}\right) \\
& =1+\frac{1}{a_{k+1}}+\frac{P_{k}}{a_{k+1}} \cdot \frac{-1}{P_{k}}=1+\frac{1}{a_{k+1}}-\frac{1}{a_{k+1}}=1
\end{aligned}
$$

By the principle of mathematical induction, it follows that $t_{n}=1$, for every $n \in \mathbb{N}$. Therefore, $S_{2009}+P_{2009}=1$.