Question

${\sum }_{m=1}^n{\tan }^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$ is equal to

• A. ${\tan }^{-1}\left(n^2+n+1\right)-\frac{\pi }{4}$
• B. ${\tan }^{-1}\left(n^2+n+1\right)+\frac{\pi }{4}$
• C. ${\tan }^{-1}\left(n^2+n-1\right)-\frac{\pi }{4}$
• D. ${\tan }^{-1}\left(n^2-n-1\right)-\frac{\pi }{4}$

Answer 1

The correct option is A ${\tan }^{-1}\left(n^2+n+1\right)-\frac{\pi }{4}$

$\sum _{m=1}^n{\tan }^{-1}\left(\frac{2m}{m^4+m^2+2}\right)$ is equal to

In General formula for $ta{n}^{-1}x=ta{n}^{-1}y$ is

$ta{n}^{-1}\left(\frac{x\pm y}{1\pm xy}\right)...\left(1\right)$

Now, given summation is,

$\sum _{m=1}^{n}ta{n}^{-1}\left(\frac{2m}{m^4+m^2+2}\right)=\sum _{m=1}^{n}ta{n}^{-1}\left(\frac{2m}{1+(m^4+m^2+1)}\right)$

Now, factorisation of. $m^4+m^2+1$ is

$m^4+m^2+1=(m^2+m+1)\times (m^2-m+1)$

and difference between $(m^2-m+1)$ and

$(m^2-m+1)$ is 2m.

$\Rightarrow \sum _{m=1}^{n}ta{n}^{-1}\left(\frac{(m^2+m+1)-(m^2-m+1)}{1+(m^2+m+1)(m^2-m+1)}\right)...\left(2\right)$

Now, compare $e{q}^n\left(1\right)$ & $e{q}^n\left(2\right)$

$=\sum _{m-1}^{n}ta{n}^{-1}(m^2+m+1)-ta{n}^{-1}(m^2\ast m+1)$

Now putting value from 1, 2, ...,n.

$=ta{n}^{-1}(m^2+1+1)-ta{n}^{-1}(1^2-1+1)+ta{n}^{-1}(2^2+2+1)$

$-ta{n}^{-1}(2^2-2+1)+...+ta{n}^{2-1}(n^2+n+1)..ta{n}^{-1}(n^2-n+1)$

$=ta{n}^{-1}\left(3\right)-ta{n}^{-1}\left(1\right)+ta{n}^{-1}\left(7\right)-ta{n}^{-1}\left(3\right)+...+ta{n}^{-1}(n^2+n+1)-ta{n}^{-1}(n^2-n+1)$

here alternative terms will canceled and we are left with $ta{n}^{-1}(n^2+n+1)-ta{n}^{-1}\left(1\right)$

So, $ta{n}^{-1}(n^2+n+1)-\frac{\pi }{4}$