6 - 16 Find the value of $10 \operatorname{ctg}(\operatorname{arcctg} 3+\operatorname{arcctg} 7+\operatorname{arcctg} 13+\operatorname{arcctg} 21)$.
[Solution] Let $a_{n}=1+n+n^{2}$, then $a_{1}=3, a_{2}=7, a_{3}=13, a_{4}=21$
We first prove a formula:
$$\operatorname{arcctg}\left(1+n+n^{2}\right)=\operatorname{arctg}(n+1)-\operatorname{arctg} n$$

In fact, we can set
$$\alpha=\operatorname{arctg}(n+1), \beta=\operatorname{arctg} n$$

Thus $0<\beta, 0<\alpha-\beta<\frac{\pi}{2}$. Also, $\operatorname{ctg}[\operatorname{arctg}(n+1)-\operatorname{arctg} n]$
$$\begin{array}{l}
=\operatorname{ctg}(\alpha-\beta) \\
=\frac{1+\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}{\operatorname{tg} \alpha-\operatorname{tg} \beta} \\
=\frac{1+(n+1) \cdot n}{(n+1)-n} \\
=1+n+n^{2}
\end{array}$$

Therefore,
$$\operatorname{arcctg}\left(1+n+n^{2}\right)=\operatorname{arctg}(n+1)-\operatorname{arctg} n$$

Let $\theta=\operatorname{arcctg} 3+\operatorname{arcctg} 7+\operatorname{arcctg} 13+\operatorname{arcctg} 21$.
By formula (1) we get
$$\begin{array}{l}
\operatorname{arcctg} 3=\operatorname{arctg} 2-\operatorname{arctg} 1 \\
\operatorname{arcctg} 7=\operatorname{arctg} 3-\operatorname{arctg} 2 \\
\operatorname{arcctg} 13=\operatorname{arctg} 4-\operatorname{arctg} 3 \\
\operatorname{arcctg} 21=\operatorname{arctg} 5-\operatorname{arctg} 4
\end{array}$$

Adding the above four equations, we get
$$\theta=\operatorname{arctg} 5-\operatorname{arctg} 1$$

Thus,
$$\begin{aligned}
& \quad 10 \operatorname{ctg}(\operatorname{arcctg} 3+\operatorname{arcctg} 7+\operatorname{arcctg} 13+\operatorname{arcctg} 21) \\
= & 10 \operatorname{ctg} \theta \\
= & 10 \operatorname{ctg}(\operatorname{arctg} 5-\operatorname{arctg} 1) \\
= & 10 \cdot \frac{1+5 \cdot 1}{5-1}=15
\end{aligned}$$