_{m=1}ⁿ ^{-1} ≤ft( {2m}{m^{4}+m²+2} ) is equal to

Correct Answer is : tan⁡−1(n2+nn2+n+2)^{-1}≤ft({n²+n}{n²+n+2})tan−1(n2+n+2n2+n)

Correct Answer is : tan⁡−1(n2+nn2+n+2)^{-1}≤ft({n²+n}{n²+n+2})tan−1(n2+n+2n2+n)

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Manipal Entrance Test 2014 Math Paper

Question : 32 of 80

Marks: +1, -0

\sum\limits_{m=1}^{n} \tan^{-1} \left( \frac{2m}{m^{4}+m^{2}+2} \right)

is equal to

Solution:

We have,

\sum\limits_{m=1}^{n} \tan^{-1} \left( \frac{2m}{m^{4}+m^{2}+2} \right)

=\sum\limits_{m=1}^{n}\tan^{-1} \left[ \frac{2m}{1+(m^{2}+m+1)(m^{2}-m+1)} \right]

=\sum\limits_{m=1}^{n} \tan^{-1} \left[ \frac{(m^{2}+m+1)-(m^{2}-m+1)}{1+(m^{2}+m+1)(m^{2}-m+1)} \right]

=\sum\limits_{m=1}^{n} \left\{ \tan^{-1} [m^{2}+m+1] - \tan^{-1} [m^{2}-m+1] \right\}

=\tan^{-1} 3 - \tan^{-1} 1 + \tan^{-1} 7 - \tan^{-1} 3 + (\tan^{-1} 13 - \tan^{-1} 7) + \cdots + \tan^{-1}(n^{2}+n+1) - \tan^{-1}(n^{2}-n+1)

=\tan^{-1}\left(\frac{n^{2}+n+1-1}{1+(n^{2}+n+1)\cdot 1}\right)

=\tan^{-1}\left(\frac{n^{2}+n}{2+n^{2}+n}\right)

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