The value of $\cot\left(\sum\limits^{19}_{n=1} \cot^{-1} \left(1+ \sum\limits^{n}_{p=1} 2p\right)\right) $ is :

Question

The Range of $f(x) = \sin^{-1} \left( \frac{1}{x^2 - 2x + 2} \right)$ is:

Answer 1

To solve the problem, we need to evaluate the expression:

$\cot\left(\sum\limits^{19}_{n=1} \cot^{-1} \left(1+ \sum\limits^{n}_{p=1} 2p\right)\right)$

Let's break this problem into simpler parts.

Consider the inner sum: $\sum\limits^{n}_{p=1} 2p$. It is the sum of the arithmetic series $2 + 4 + 6 + \ldots + 2n$.

The sum of the first $n$ even numbers is given by: $\sum\limits_{p=1}^{n} 2p = 2(1 + 2 + \ldots + n) = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$.
Substituting in the main expression, we have: $\sum\limits^{19}_{n=1} \cot^{-1} \left(1 + n(n+1)\right)$.

This simplifies to: $\sum\limits^{19}_{n=1} \cot^{-1} \left(n^2 + n + 1\right)$.
We use the identity: $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$ to rephrase the sum as: $\sum\limits_{n=1}^{19} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$.

We are looking for a telescopic nature in this summation. Consider the identity: $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$.
Recognize that: $\frac{1}{n^2 + n + 1}$ can be decomposed using partial fractions or such that it will telescope.

Each term effectively cancels one from the other series term in a telescopic series. The result is simplified through: $\cot^{-1}(1) - \cot^{-1}(n^2 + n + 1)$, evaluating at the bounds.
For the sum to telescope cleanly: $\cot^{-1}(1) = \tan^{-1}(1) = \frac{\pi}{4}$ only if starting and ending terms give a clean bound difference.

Ultimately this reduces to: $\tan^{-1}\left(\frac{21}{19}\right)$.

Thus, the value of: $\cot\left(\sum\limits^{19}_{n=1} \cot^{-1} \left(n^2 + n + 1\right)\right)$ is $\frac{21}{19}$.

Finally, based on the above calculations, the correct answer is: $\frac{21}{19}$.

Answer 2

To solve the problem, we need to evaluate the expression:

$\cot\left(\sum\limits^{19}_{n=1} \cot^{-1} \left(1+ \sum\limits^{n}_{p=1} 2p\right)\right)$

Let's break this problem into simpler parts.

Consider the inner sum: $\sum\limits^{n}_{p=1} 2p$. It is the sum of the arithmetic series $2 + 4 + 6 + \ldots + 2n$.

The sum of the first $n$ even numbers is given by: $\sum\limits_{p=1}^{n} 2p = 2(1 + 2 + \ldots + n) = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$.
Substituting in the main expression, we have: $\sum\limits^{19}_{n=1} \cot^{-1} \left(1 + n(n+1)\right)$.

This simplifies to: $\sum\limits^{19}_{n=1} \cot^{-1} \left(n^2 + n + 1\right)$.
We use the identity: $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$ to rephrase the sum as: $\sum\limits_{n=1}^{19} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$.

We are looking for a telescopic nature in this summation. Consider the identity: $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$.
Recognize that: $\frac{1}{n^2 + n + 1}$ can be decomposed using partial fractions or such that it will telescope.

Each term effectively cancels one from the other series term in a telescopic series. The result is simplified through: $\cot^{-1}(1) - \cot^{-1}(n^2 + n + 1)$, evaluating at the bounds.
For the sum to telescope cleanly: $\cot^{-1}(1) = \tan^{-1}(1) = \frac{\pi}{4}$ only if starting and ending terms give a clean bound difference.

Ultimately this reduces to: $\tan^{-1}\left(\frac{21}{19}\right)$.

Thus, the value of: $\cot\left(\sum\limits^{19}_{n=1} \cot^{-1} \left(n^2 + n + 1\right)\right)$ is $\frac{21}{19}$.

Finally, based on the above calculations, the correct answer is: $\frac{21}{19}$.

The Correct Option is C