A Nice Problem from JEE Mains 2019 which I got wrong:(

Let $\displaystyle X_k = \cot \left(\sum_{n=1}^k \cot^{-1} \left(1+\sum_{p=1}^n 2p \right) \right) = \cot \left(\sum_{n=1}^k \cot^{-1} \left(n^2+n+1 \right) \right)$ . From the first few $k$ 's, $X_1 = 3 = \frac 31$ , $X_2 = 2 = \frac 42$ , $X_3 = \frac 53$ , $X_4 = \frac 32 = \frac 64$ ..., we note that $X_k = \dfrac {k+2}k$ . Let us prove by induction that this claim is true for all $k \ge 1$ .

For $k=1$ , $X_1 = \cot \left(\cot^{-1} \left(1^2+1+1 \right) \right) = 3 = \dfrac {1+2}1$ . Therefore the claim is true for $k=1$ .

Assuming the claim is true for $k$ , then

$\begin{aligned} X_{\color{#D61F06}k+1} & = \cot \left(\sum_{n=1}^{\color{#D61F06}k+1} \cot^{-1} \left(n^2+n+1 \right) \right) \\ & = \cot \left(\sum_{n=1}^{\color{#3D99F6}k} \cot^{-1} \left(n^2+n+1 \right) + \cot^{-1}\left(({\color{#D61F06}k+1})^2 + ({\color{#D61F06}k+1})+1\right) \right) \\ & = \cot \left( \cot^{-1} \left(\color{#3D99F6} \frac {k+2}k \right) + \cot^{-1}(k^2 + 3k+3) \right) \\ & = \frac {\left(\frac {k+2}k\right)(k^2 + 3k+3)-1}{\frac {k+2}k + k^2 + 3k+3} \\ & = \frac {k^3 + 5k^2 + 8k+6}{k^3 + 3k^2 + 4k+2} \\ & = \frac {(k+3)(k^2+2k+2)}{(k+1)(k^2+2k+2)} \\ & = \frac {(k+1)+2}{k+1} \end{aligned}$

Therefore the claim is also true for $k+1$ and hence true for all $k \ge 1$ . And $X = X_{19} = \dfrac {19+2}{19} = \boxed{\dfrac {21}{19}}$ .