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

Question

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

I tried to simplify it but it became messy. How can I find $\sum_{n=1}^{23} \cot^{-1} (1+n+n^2)$? Please help me.

Answer 1

The hint.

Use $$\arctan(n+1)-\arctan{n}=arccot(n^2+n+1)$$ and the telescopic summation.