Proving $\arctan1 + \arctan\frac13 +\cdots+\arctan\frac1{n^2+n+1}=\arctan (n+1)$ by induction

Question

How to solve this problem using mathematical induction? $$\arctan1 + \arctan\frac13 +\cdots+\arctan\frac1{n^2+n+1}=\arctan (n+1)$$

Answer 1

Hint: $$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$

In your case, for the base case $n=1$, $$\tan\left(\arctan 1+\arctan\frac{1}{3}\right)=\frac{\tan(\arctan 1)+\tan(\arctan\frac{1}{3})}{1-\tan(\arctan1)\tan(\arctan\frac{1}{3})}=\frac{1+\frac{1}{3}}{1-\frac{1}{3}}=2$$So $\arctan 1+\arctan\frac{1}{3}=\arctan 2$.