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Step-by-Step Solution
Step 1: Understand the Problem
We are given a sequence of consecutive natural numbers:
$a_1 = 1, a_2 = 2, a_3 = 3,\ldots, a_{2022} = 2022$. We want to find the value of the sum
$$
\tan^{-1}\left(\frac{1}{1 + a_1 a_2}\right)
+ \tan^{-1}\left(\frac{1}{1 + a_2 a_3}\right)
+ \cdots +
\tan^{-1}\left(\frac{1}{1 + a_{2021}a_{2022}}\right).
$$
Step 2: Express Each Term Using a Known Identity
A useful identity for differences of inverse tangents is:
$$
\tan^{-1}(x) - \tan^{-1}(y)
= \tan^{-1}\!\Bigl(\frac{x-y}{1 + x\,y}\Bigr).
$$
From this, one can derive that:
$$
\tan^{-1}\!\Bigl(\frac{1}{1 + a\,b}\Bigr)
= \tan^{-1}\!\Bigl(\frac{1}{a}\Bigr)
- \tan^{-1}\!\Bigl(\frac{1}{b}\Bigr),
$$
when $b = a + 1$ (for consecutive integers).
Step 3: Apply the Identity to Each Term
Since $a_{1} = 1, a_{2} = 2, a_{3}=3, \ldots, a_{2022}=2022$, each term in the sum can be written as:
$$
\tan^{-1}\left(\frac{1}{1 + a_{k}\,a_{k+1}}\right)
= \tan^{-1}\!\Bigl(\frac{1}{a_{k}}\Bigr)
- \tan^{-1}\!\Bigl(\frac{1}{a_{k+1}}\Bigr),
$$
for $k = 1, 2, \ldots, 2021$.
Step 4: Observe the Telescoping Pattern
Notice that when you add these terms in sequence, most of the intermediate terms cancel out (telescoping series). Specifically:
$$
\bigl[\tan^{-1}(\tfrac{1}{1}) - \tan^{-1}(\tfrac{1}{2})\bigr]
+ \bigl[\tan^{-1}(\tfrac{1}{2}) - \tan^{-1}(\tfrac{1}{3})\bigr]
+ \cdots
+ \bigl[\tan^{-1}(\tfrac{1}{2021}) - \tan^{-1}(\tfrac{1}{2022})\bigr].
$$
Almost everything in the middle cancels, leaving:
$$
= \tan^{-1}(1) - \tan^{-1}\!\Bigl(\frac{1}{2022}\Bigr).
$$
Step 5: Simplify the Final Expression
We know that $\tan^{-1}(1) = \frac{\pi}{4}$. Also,
$$
\tan^{-1}\!\Bigl(\frac{1}{2022}\Bigr)
= \frac{\pi}{2} - \cot^{-1}(2022)
= \frac{\pi}{2} - \bigl(\frac{\pi}{2} - \tan^{-1}(2022)\bigr)
= \tan^{-1}(2022).
$$
Hence,
$$
\tan^{-1}(1) - \tan^{-1}\!\Bigl(\frac{1}{2022}\Bigr)
= \frac{\pi}{4} - \bigl(\frac{\pi}{2} - \tan^{-1}(2022)\bigr)
= \tan^{-1}(2022) - \frac{\pi}{4}.
$$
Final Answer
Therefore, the given sum equals
$$
\tan^{-1}(2022) - \frac{\pi}{4}.
$$
