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:
$$
\operatorname{arcctg}\left(1+n+n^{2}\right)=\operatorname{arctg}(n+1)-\operatorname{arctg} n \text {. }
$$

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

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

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

Let $\quad \theta=\operatorname{arcctg} 3+\operatorname{arcctg} 7+\operatorname{arcctg} 13+\operatorname{arcctg} 21$.
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
$$
\theta=\operatorname{arctg} 5-\operatorname{arctg} 1 \text {. }
$$

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