Let a_1=1,a_2,a_3,a_4, ..

Question

Let ${a}_{1}=1,{a}_{2},{a}_{3},{a}_{4},\ldots .$. be consecutive natural numbers.

Then ${	an }^{-1}(\frac{1}{1+{a}_{1}{a}_{2}})+{	an }^{-1}(\frac{1}{1+{a}_{2}{a}_{3}})+\ldots ..+{	an }^{-1}(\frac{1}{1+{a}_{2021}{a}_{2022}})$ is equal to :

Accepted Answer

$a_{1}=1, a_{2}, a_{3}, a_{4}, \ldots .$. be consecutive natural numbers.

$$
\begin{aligned}
& \therefore \quad a_2=2, a_3=3, \ldots ., a_{2021}=2021, a_{2022}=2022 \\
& \tan ^{-1}\left[\frac{1}{1+a_1 a_2}\right]=\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2}\right) \\
& \tan ^{-1}\left[\frac{1}{1+a_2 a_3}\right]=\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]=\tan ^{-1}\left(\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{3}\right)
\end{aligned}
$$

.

$$
\begin{aligned}
\tan ^{-1}\left[\frac{1}{1+a_{2021} a_{2022}}\right] & =\tan ^{-1}\left[\frac{1}{1+2021 \cdot 2022}\right] \\
& =\tan ^{-1}\left(\frac{1}{2021}\right)-\tan ^{-1}\left(\frac{1}{2022}\right)
\end{aligned}
$$

$$ \therefore $$ $\tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots . .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$

$$
=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{2022}\right)=\frac{\pi}{4}-\cot ^{-1}(2022)
$$

$$
=\frac{\pi}{4}-\left(\frac{\pi}{2}-\tan ^{-1}(2022)\right)=\tan ^{-1}(2022)-\frac{\pi}{4}
$$

Solutions · 1