JEE Advanced 2013 · Paper 1, Mathematics question 47

Inner sum. $1+\sum_{k=1}^n2k=1+n(n+1)$.

Telescoping. $\cot^{-1}(1+n(n+1))=\tan^{-1}\dfrac{(n+1)-n}{1+n(n+1)}=\tan^{-1}(n+1)-\tan^{-1}n$, so the sum is $\tan^{-1}24-\tan^{-1}1=\tan^{-1}\dfrac{23}{25}$.

Value. $\cot\left(\tan^{-1}\frac{23}{25}\right)=\dfrac{25}{23}$.

(B)