2255 $\underset{\star}{\star \star}$ Find the value of $10 \cot (\operatorname{arccot} 3+\operatorname{arccot} 7+\operatorname{arccot} 13+\operatorname{arccot} 21)$.
Let $a_{n}=1+n+n^{2}$, then $a_{1}=3, a_{2}=7, a_{3}=13, a_{4}=21$. We will first prove a formula:
$$
\operatorname{arccot}\left(1+n+n^{2}\right)=\arctan (n+1)-\arctan n
$$

In fact, let $\alpha=\arctan (n+1), \beta=\arctan n$. Then $0<\beta<\alpha$, $0<\alpha-\beta<\frac{\pi}{2}$. Also,
$$
\begin{aligned}
& \cot [\arctan (n+1)-\arctan n]=\cot (\alpha-\beta) \\
= & \frac{1+\tan \alpha \cdot \tan \beta}{\tan \alpha-\tan \beta}=\frac{1+(n+1) \cdot n}{(n+1)-n} \\
= & 1+n+n^{2} .
\end{aligned}
$$

Therefore,
$$
\operatorname{arccot}\left(1+n+n^{2}\right)=\arctan (n+1)-\arctan n \text {. }
$$

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

Adding the above four equations, we get $\theta=\arctan 5-\arctan 1$. Thus,
$$
\begin{aligned}
& 10 \cot (\operatorname{arccot} 3+\operatorname{arccot} 7+\operatorname{arccot} 13+\operatorname{arccot} 21) \\
= & 10 \cot \theta=10 \cot (\arctan 5-\arctan 1) \\
= & 10 \cdot \frac{1+5 \cdot 1}{5-1}=15 .
\end{aligned}
$$