Let $a_1=1, a_2, a_3, a_4, \ldots $ be consecutive natural numbersThen $\tan ^{-1}\left(\frac{1}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{1}{1+a_2 a_3}\right)+\ldots +\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$ is equal to

Question

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$, then $\displaystyle\sum_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2$ is equal to

Answer 1

To solve the problem, we need to simplify the sum:

$\tan^{-1}\left(\frac{1}{1 + a_1 a_2}\right) + \tan^{-1}\left(\frac{1}{1 + a_2 a_3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1 + a_{2021} a_{2022}}\right)$

where $a_1 = 1$, and the sequence $a_n = n$ for natural numbers.

Consider the expression $\tan^{-1}\left(\frac{1}{1 + n(n+1)}\right)$. We can use the identity:

$\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x + y}{1 - xy}\right)$

If $x = \frac{1}{1 + n(n+1)}$, then according to the identity, let's examine:

$\tan^{-1} x = \tan^{-1} \left( \frac{1}{2n+1} \right)$. This transforms using the fact $\frac{1}{a + x(a)}$ leads to a chain telescoping effect.

Understanding this, we observe a pattern similar to:

$\tan^{-1}(n+1) - \tan^{-1}(n)$.

Notice the telescopic nature of this expression across the large series. It implies that all intermediate terms cancel each other.

From the pattern of the telescopic series, the expression collapses to:

$\tan^{-1}(2022) - \tan^{-1}(1)$

However, since:

$\tan^{-1}(1) = \frac{\pi}{4}$

We are left with:

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

The equivalent to this modification using inverse tangent and cotangent property can be altered as:

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

Hence the correct answer is:

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

Answer 2