Let a₁ = 1, a₂, a₃, a₄, ... be consecutive natural numbers. Then tan⁻¹(1/(1+a₁a₂)) + tan⁻¹(1/(1+a₂a₃

Q: Wait, why do we reject the negative root here? Or is there no negative root?

In this specific problem, we are dealing with natural numbers $a_n = n$. Since $n$ is always positive, all our terms are well-defined and positive. There are no quadratic equations here, so no roots to reject! We are simply using a trigonometric identity to simplify a sum.

Q: How did you know to replace the '1' with $a_{n+1} - a_n$?

That is the 'Aha!' moment of the problem. In series problems involving $\tan^{-1}$, we almost always look for a way to create the form $\frac{x-y}{1+xy}$. Since we had $1+a_n a_{n+1}$ in the denominator, we needed the numerator to be the difference of $a_{n+1}$ and $a_n$. Recognizing that $a_{n+1} - a_n = 1$ is the key to unlocking the telescoping pattern.

Q: Why can't we just calculate each term individually?

You could, but you would be there until the next century! The sum goes up to $2021$ terms. JEE problems are designed to test your ability to spot patterns and use identities to bypass brute-force calculation. Telescoping series are a classic tool for this.

Question

Let $a_1 = 1, a_2, a_3, a_4, \dots$ be consecutive natural numbers. Then $	an^{-1}\left(\frac{1}{1+a_1 a_2}ight) + 	an^{-1}\left(\frac{1}{1+a_2 a_3}ight) + \dots + 	an^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}ight)$ is equal to

Options:

(A) $\frac{\pi}{4} - \cot^{-1}(2022)$

(B) $\cot^{-1}(2022) - \frac{\pi}{4}$

(C) $	an^{-1}(2022) - \frac{\pi}{4}$

(D) $\frac{\pi}{4} - 	an^{-1}(2022)$

Sigma Percentile · Accepted Answer

### The Beauty of Telescoping Series

Imagine you are standing before a massive, intimidating sum:
$$\sum_{n=1}^{2021} 	an^{-1}\left(\frac{1}{1+a_n a_{n+1}}ight)$$

At first glance, it looks like a nightmare of calculation. But in the world of JEE Advanced, intimidation is just a mask for elegance. Let us peel back that mask together.

### Phase 1: Decoding the Sequence

The problem provides a sequence of consecutive natural numbers starting with $a_1 = 1$. This is our foundation.

If $a_1 = 1$, then $a_2 = 2$, $a_3 = 3$, and generally, $a_n = n$. This simple realization is our first step toward victory.

The difference between any two adjacent terms is constant: $a_{n+1} - a_n = 1$. Keep this in your pocket; it is the master key.

### Phase 2: The Strategic Manipulation

Look at the general term of our series:
$$T_n = 	an^{-1}\left(\frac{1}{1 + a_n a_{n+1}}ight)$$

That '1' in the numerator is suspicious. It is too perfect. We know that $a_{n+1} - a_n = 1$.

Let us perform a substitution. Replace that '1' with $(a_{n+1} - a_n)$. Now, our term looks like this:
$$T_n = 	an^{-1}\left(\frac{a_{n+1} - a_n}{1 + a_n a_{n+1}}ight)$$

Do you see it? It is the exact form of the inverse tangent subtraction identity:
$$	an^{-1}\left(\frac{x-y}{1+xy}ight) = 	an^{-1}(x) - 	an^{-1}(y)$$

### Phase 3: The Telescoping Magic

With the identity in hand, our complex term $T_n$ collapses into something beautiful:
$$T_n = 	an^{-1}(a_{n+1}) - 	an^{-1}(a_n)$$

Now, let us write out the sum. For $n=1$, we have $	an^{-1}(a_2) - 	an^{-1}(a_1)$. For $n=2$, we have $	an^{-1}(a_3) - 	an^{-1}(a_2)$.

As we continue this, notice the pattern: the positive part of one term cancels the negative part of the next. This is the 'telescoping' effect—like a collapsible telescope folding into itself.

All the intermediate terms vanish, leaving only the very first negative part and the very last positive part:
$$S = 	an^{-1}(a_{2022}) - 	an^{-1}(a_1)$$

### Phase 4: The Final Victory

We are almost there. We know $a_{2022} = 2022$ and $a_1 = 1$. Substituting these values, we get:
$$S = 	an^{-1}(2022) - 	an^{-1}(1)$$

Since $	an^{-1}(1) = \frac{\pi}{4}$, our final result is:
$$\mathbf{	an^{-1}(2022) - \frac{\pi}{4}}$$

It is elegant, it is precise, and it is the result of seeing the structure beneath the numbers. You have successfully navigated the trap and found the path to the solution. Keep this mindset—always look for the pattern, always trust the identity.

Select Answer:

Visualized Solution

Identifying the Sequence $a_{n}$

The Common Difference

Analyzing the General Term $T_{n}$

Creating the Difference in Numerator

Applying the Inverse Tangent Identity

Splitting the General Term

Expanding the First Few Terms

Expanding up to the Last Term

The Telescoping Effect

The Surviving Terms

Final Substitution and Result

The Sigma Insight: Properties of Inverse Trigonometric Functions

The Beauty of Telescoping Series

Phase 1

Phase 2

Phase 3

Phase 4

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