I am given $$S=\sum\limits_{n=1}^{23}\cot^{-1}\left(1+ \sum\limits_{k=1}^n 2k\right)$$
On expanding the sigma series becomes $$S= 23\cot^{-1}(3)+22\cot^{-1}(5) + \cdots + \cot^{-1}(47)$$ And in tan form as $$S= 23\tan^{-1}(1/3)+22\tan^{-1}(1/5) + \cdots + \tan^{-1}(1/47)$$ How to sum this series?
$$S=\sum\limits_{n=1}^{23}\cot^{-1}\left(1+ \sum\limits_{k=1}^n 2k\right)=\sum_{n=1}^{23}\cot^{-1}(1+n(n+1))=\sum_{n=1}^{23}\arctan\left(\frac{(n+1)-n}{1+n(n+1)}\right)=\sum_{n=1}^{23}[\arctan (n+1)-\arctan (n)]=\arctan(24)-\arctan(1)=\arctan\left(\frac{23}{25}\right)$$