The value of cot (Σ_{n=1}^{23} cot⁻¹ (1 + Σ_{k=1}^n 2k)) is (A) 23/25 (B) 25/23 (C) 23/24 (D) 24/23

Q: Wait, why do we convert $\cot^{-1}$ to $\tan^{-1}$? Can't we just solve it with $\cot^{-1}$?

While you technically could, it is much harder! The identity for the difference of two $\cot^{-1}$ terms is messy. By converting to $\tan^{-1}$, we unlock the elegant $\tan^{-1}(x) - \tan^{-1}(y)$ identity, which is the standard tool for solving telescoping series in JEE problems.

Q: How do I know if the series is telescoping?

Whenever you see a general term $T_n$ that can be expressed as $f(n+1) - f(n)$, you are looking at a telescoping series. In this problem, after the transformation, we get $\tan^{-1}(n+1) - \tan^{-1}(n)$. When you sum these up, the intermediate terms cancel out, leaving only the boundaries.

Q: Why does the final answer flip from $23/25$ to $25/23$?

Great catch! We found that the summation equals $\tan^{-1}(\frac{23}{25})$. The original expression asks for the $\cot$ of this sum. Since $\cot(\tan^{-1}(x)) = \frac{1}{x}$, the $\cot$ of $\tan^{-1}(\frac{23}{25})$ is simply the reciprocal, $\frac{25}{23}$.

Question

The value of $\cot \left( \sum_{n=1}^{23} \cot^{-1} \left( 1 + \sum_{k=1}^n 2k \right) \right)$ is

Options:

(A) 23/25

(B) 25/23

(C) 23/24

(D) 24/23

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, multi-layered puzzle. At first glance, the expression

$$\cot \left( \sum_{n=1}^{23} \cot^{-1} \left( 1 + \sum_{k=1}^n 2k ight) ight)$$

looks like a labyrinth. In the world of JEE Advanced, complexity is often just a mask for elegance. We are going to peel back these layers one by one.

### Unraveling the Inner Core

Let us begin at the very heart of the expression: the innermost summation, $\sum_{k=1}^n 2k$. This is a simple arithmetic progression.

We can factor out the constant $2$, leaving us with $2 \sum_{k=1}^n k$. Using the standard formula for the sum of the first $n$ natural numbers,

$$\frac{n(n+1)}{2}$$

our expression simplifies beautifully: $2 \cdot \frac{n(n+1)}{2} = n(n+1)$. Just like that, the intimidating inner sum collapses into a clean, quadratic form.

### The Transformation

Now, we substitute this back into our main expression. The term inside the $\cot^{-1}$ becomes $1 + n(n+1)$.

We are now looking at $\cot^{-1}(1 + n(n+1))$. Working with $\cot^{-1}$ is rarely the path of least resistance, so we use the identity $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ to transform our term into:

$$	an^{-1} \left( \frac{1}{1 + n(n+1)} ight)$$

This is the "Aha!" moment. The structure $\frac{1}{1 + n(n+1)}$ is a classic setup for the method of differences.

### The Telescoping Cascade

To unlock the telescoping series, we need to express the numerator $1$ as the difference of the two factors in the denominator. We rewrite $1$ as $(n+1) - n$.

Now our term looks like this:

$$T_n = 	an^{-1} \left( \frac{(n+1) - n}{1 + (n+1)n} ight)$$

This perfectly matches the trigonometric identity $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$. Thus, our general term becomes $T_n = 	an^{-1}(n+1) - 	an^{-1}(n)$.

When we sum this from $n=1$ to $23$, we get:

$$(	an^{-1}(2) - 	an^{-1}(1)) + (	an^{-1}(3) - 	an^{-1}(2)) + \dots + (	an^{-1}(24) - 	an^{-1}(23))$$

Notice the pattern? The positive $	an^{-1}(2)$ cancels the negative $	an^{-1}(2)$, and so on. Only the first negative term and the last positive term survive: $	an^{-1}(24) - 	an^{-1}(1)$.

### The Final Victory

We are almost there. We have the sum as $	an^{-1}(24) - 	an^{-1}(1)$. Applying the subtraction formula again, we get:

$$	an^{-1} \left( \frac{24-1}{1 + 24 \cdot 1} ight) = 	an^{-1} \left( \frac{23}{25} ight)$$

Finally, we return to the outermost function: $\cot \left( 	an^{-1} \left( \frac{23}{25} ight) ight)$. Since $\cot(	an^{-1}(x)) = \frac{1}{x}$, our final answer is:

$$\mathbf{\frac{25}{23}}$$

It is a journey from chaos to order, and that is the essence of mathematics.

Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of $cot (\sum_{n = 1}^{23} cot^{- 1} (1 + \sum_{k = 1}^{n} 2 k))$ is

Select Answer:

Visualized Solution

Expression Analysis

Inner Summation $\sum 2 k$

Substitution and Conversion

Method of Differences

Applying Identity

Telescoping Cancellation

Simplifying the Sum

Final Calculation

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

Unraveling the Inner Core

The Transformation

The Telescoping Cascade

The Final Victory

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