Question

If ${\sum }_{r=1}^{10}ta{n}^{-1}\left[\frac{3}{9r^2+3r-1}\right]=co{t}^{-1}\left[\frac{m}{n}\right]$ (where m and n are coprime), then find the value of (m - n). ___

Answer 1

$a_n=ta{n}^{-1}\left[\frac{3}{9n^2+3n-1}\right]$
$=ta{n}^{-1}\left[\frac{3}{1+(3n+2)(3n-1)}\right]\Rightarrow ta{n}^{-1}\left[\frac{(3n+2)-(3n-1)}{1+(3n+2)(3n-1)}\right]$
$=ta{n}^{-1}(3n+2)-ta{n}^{-1}(3n-1)$
$\therefore$ Sum of first 10 terms
$={\sum }_{r=1}^{10}{a}_r={\sum }_{r=1}^{10}\left[\right(ta{n}^{-1}(3r+2)-ta{n}^{-1}(3r-1)\left)\right]$
$=(ta{n}^{-1}32-ta{n}^{-1}2)=ta{n}^{-1}\left(\frac{32-2}{1+32.2}\right)$
$=ta{n}^{-1}\frac{30}{65}=ta{n}^{-1}\left(\frac{6}{13}\right)=co{t}^{-1}\left(\frac{13}{6}\right)=co{t}^{-1}\left(\frac{m}{n}\right)$
$\therefore$ m=13 and n = 6
Hence (m-n)= 13-6=7