Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>Let $y = \cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3) + \cdots$. If $\dfrac{d}{dn}\!\left[\sum_{r=1}^{n}\cot^{-1}(r)\right]$ is evaluated and the sum $\sum_{r=1}^{10}\cot^{-1}(r^2-r+1)$ equals $\tan^{-1}(k/l)$, find $10k+l$ (answer 101 from key).</p>
Step-by-Step Solution
Key Concept: General
<b>Telescoping Sum of Inverse Cotangents</b><br>
Key identity: $\cot^{-1}(r^2-r+1) = \tan^{-1}\!\left(\dfrac{1}{r^2-r+1}\right) = \tan^{-1}(r) - \tan^{-1}(r-1)$.<br>
This is because: $\tan^{-1}(r)-\tan^{-1}(r-1) = \tan^{-1}\!\left(\dfrac{r-(r-1)}{1+r(r-1)}\right) = \tan^{-1}\!\left(\dfrac{1}{r^2-r+1}\right)$.<br>
$\sum_{r=1}^{10}\cot^{-1}(r^2-r+1) = \sum_{r=1}^{10}[\tan^{-1}(r)-\tan^{-1}(r-1)] = \tan^{-1}(10)-\tan^{-1}(0) = \tan^{-1}(10)$.<br>
If the answer is expressed as $\tan^{-1}(10/1)$, then $k=10$, $l=1$, $10k+l=101$.<br>
<b>Answer: 101</b><br>
<b>Key concept:</b> Telescoping with inverse trig: $\tan^{-1}A-\tan^{-1}B = \tan^{-1}\dfrac{A-B}{1+AB}$.<br>
<b>Trap:</b> Not recognising the telescoping structure; computing each term separately.
Correct Answer: 101