Prove that : `sum_(m=1)^n\ \ \ tan^(-1)((2m)/(m^4+m^2+2))=tan^(-1)((n^2+n+1)-(pi)/4)`

Explanation

`sum_{{m}=1}^{{n}} tan ^{-1}(frac{2 {~m}}{{~m}^{4}+{m}^{2}+2})` is equal to

As we know

`tan ^{-1}(frac{x pm y}{1 pm x y}) ldots(1)`

So

`sum_{{m}=1}^{{n}} tan ^{-1}(frac{2 {~m}}{{~m}^{4}+{m}^{2}+2})=sum_{{m}=1}^{{n}} tan ^{-1}(frac{2 {~m}}{1+({m}^{4}+{m}^{2}+1)})`

factorisation of. `{m}^{4}+{m}^{2}+1` is

`{m}^{4}+{m}^{2}+1=({m}^{2}+{m}+1) times({m}^{2}-{m}+1)`

difference between `({m}^{2}-{m}+1)` and

...