14. Let $x$ be the smaller of the two solutions of the equation $x^{2}-4 x+2=0$. What are the first three digits after the decimal point in the base 10 representation of the number

$$
x+x^{2}+x^{3}+\cdots+x^{2009} ?
$$
14. The answer is 414. From the usual solution formula for quadratic equations, we have $x=2-\sqrt{2}$. Using now the formula for the sum of a geometric progression, we have

$$
\begin{aligned}
x+x^{2}+x^{3}+\cdots+x^{2009} & =x\left(1+x+x^{2}+\cdots+x^{2008}\right) \\
& =x \frac{1-x^{2009}}{1-x}=\frac{x}{1-x}-\frac{x^{2010}}{1-x} .
\end{aligned}
$$

The second term of this sum is very small: indeed, with very rough estimates, we get

$$
(2-\sqrt{2})^{2010}<(0.6)^{2010}<\left((0.6)^{2}\right)^{1005}<\frac{1}{2^{1005}}<\frac{1}{\left(2^{10}\right)^{100}}<\frac{1}{10^{100}} .
$$

The first term, on the other hand, is

$$
\frac{x}{1-x}=\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{(2-\sqrt{2})(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{\sqrt{2}}{1} .
$$

Therefore, the value of the expression is a quantity that differs from $\sqrt{2}$ by a number less than $10^{-100}$; the first digits of its decimal expansion will therefore be the same as those of $\sqrt{2}$, i.e., $1.414 \ldots$.