nice exercise about arthematic sequence for \(a_{n}\) using \(\arctan\) series asked in math group

Exercise:

Let ${{\left( {{a}_{n}} \right)}_{n}}$ be an arithmetic sequence of common difference $d$

Show that,

        $\tan \left[ \arctan \left( \frac{d}{1+{{a}_{1}}{{a}_{2}}} \right)+\arctan \left( \frac{d}{1+{{a}_{2}}{{a}_{3}}} \right)+...+\arctan \left( \frac{d}{1+{{a}_{n-1}}{{a}_{n}}} \right) \right]=\frac{{{a}_{n}}-{{a}_{1}}}{1+{{a}_{1}}{{a}_{n}}}$

Solution: We have also ${{\left( {{a}_{n}} \right)}_{n}}$ is an arithmetic sequence of common difference $d$

i.e $d={{a}_{n}}-{{a}_{n-1}}$ for $n=1,2,3,....$

so $\tan \left[ \arctan \left( \frac{{{a}_{2}}-{{a}_{1}}}{1+{{a}_{1}}{{a}_{2}}} \right)+\arctan \left( \frac{{{a}_{3}}-{{a}_{1}}}{1+{{a}_{2}}{{a}_{3}}} \right)+...+\arctan \left( \frac{{{a}_{n}}-{{a}_{n-1}}}{1+{{a}_{n-1}}{{a}_{n}}} \right) \right]$

by using the fact $\arctan \left( \frac{u-v}{1+uv} \right)=\arctan \left( u \right)-\arctan \left( v \right)$

Let $x=\arctan \left( u \right)\,\,\,\And \,\,\,y=\arctan \left( v \right)\Leftrightarrow \tan \left( x-y \right)=\frac{\tan \left( x \right)-\tan \left( y \right)}{1+\tan \left( x \right)\tan \left( y \right)}$

$\Leftrightarrow x-y=\arctan \left( \frac{u-v}{1+uv} \right)\Leftrightarrow \arctan \left( u \right)-\arctan \left( v \right)=\arctan \left( \frac{u-v}{1+uv} \right)$

So $\tan \left[ \arctan \left( {{a}_{2}} \right)-\arctan \left( {{a}_{1}} \right)+\arctan \left( {{a}_{3}} \right)-\arctan \left( {{a}_{2}} \right)+....+\arctan \left( {{a}_{n}} \right)-\arctan \left( {{a}_{n-1}} \right) \right]$

$=\tan \left[ \arctan \left( {{a}_{n}} \right)-\arctan \left( {{a}_{1}} \right) \right]=\tan \left( \arctan \left( \frac{{{a}_{n}}-{{a}_{1}}}{1+{{a}_{1}}{{a}_{n}}} \right) \right)=\frac{{{a}_{n}}-{{a}_{1}}}{1+{{a}_{1}}{{a}_{n}}}$