Question

If $\displaystyle\sum_{r=1}^{10} \tan ^{-1}\left(\frac{3}{9 r^2+3 r-1}\right)=\cot ^{-1}\left(\frac{m}{n}\right)$ (where $m$ and $n$ are coprime), then find the value of $( m - n )$

Answer 1

We have
$a_n =\tan ^{-1}\left(\frac{3}{9 n^2+3 n-1}\right)=\tan ^{-1}\left(\frac{3}{1+(3 n+2)(3 n-1)}\right)=\tan ^{-1}\left(\frac{(3 n+2)-(3 n-1)}{1+(3 n+2)(3 n-1)}\right) $
$ =\tan ^{-1}(3 n+2)-\tan ^{-1}(3 n-1)$
$\therefore \text { Sum of first } 10 \text { terms }=\displaystyle\sum_{r=1}^{10} a_r=\displaystyle\sum_{r=1}^{10}\left(\tan ^{-1}(3 r+2)-\tan ^{-1}(3 r-1)\right) $
$=\left(\tan ^{-1} 5-\tan ^{-1} 2\right)+\left(\tan ^{-1} 8-\tan ^{-1} 5\right)+\ldots \ldots . .+\left(\tan ^{-1} 32-\tan ^{-1} 29\right) $
$=\tan ^{-1} 32-\tan ^{-1} 2=\tan ^{-1}\left(\frac{32-2}{1+32 \cdot 2}\right)=\tan ^{-1} \frac{30}{65}=\tan ^{-1}\left(\frac{6}{13}\right)=\cot ^{-1}\left(\frac{13}{6}\right)=\cot ^{-1}\left(\frac{ m }{ n }\right) $
$\therefore m =13 \text { and } n =6$
$\text { Hence }( m - n )=13-6=7 $