Prove the following-
\[\sum\limits_{\lambda = 1}^n {{{\tan }^{ - 1}}\left( {\dfrac{{2\lambda }}{{1 + {\lambda ^2} + {\lambda ^4}}}} \right)} = {\tan ^{ - 1}}\left( {{n^2} + n + 1} \right) - \dfrac{\pi }{4}\]

Hint: In This question, we will use the properties of inverse trigonometric functions. The formula is given by-
\[{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x - y}}{{1 + xy}}} \right)\]
We will bring the given summation in this form of tan-1x - tan-1y so that the successive terms get cancelled and only one term is left.

Complete step by step answer:
Now we can write that-
\[1 + {\lambda ^2} + {\lambda ^4}\]
\[ = {\lambda ^4} + 2{\lambda ^2} + 1 + 1 - {\lambda ^2}\]
\[ = {\left( {{\lambda ^2} + 1} \right)^2} - {\lambda ^2} + 1\]
Using the property a2 - b2 = (a + b)(a - b),
\[ = \left( {{\lambda ^2} + \lambda + 1} \right)\left( {{\lambda ^2} - \lambda + 1} \right) + 1\]
Now we can write the summation as-
\[\mathop \sum \limits_{\lambda = 1}^n {\tan ^{ - 1}}\left( {\dfrac{{\left( {{\lambda ^2} + \lambda + 1} \right) - \left( {{\lambda ^2} - \lambda + 1} \right)}}{{1 + \left( {{\lambda ^2} + \lambda + 1} \right)\left( {{\lambda ^2} - \lambda + 1} \right)}}} \right)\]
Using the formula,
\[\mathop \sum \limits_{\lambda = 1}^n \left[ {{{\tan }^{ - 1}}\left( {{\lambda ^2} + \lambda + 1} \right) - {{\tan }^{ - 1}}\left( {{\lambda ^2} - \lambda + 1} \right)} \right]\]
On substituting the values,
= tan-1(3) - tan-1(1) + tan-1(7) - tan-1(3) + ….. + tan-1(n2 + n + 1)
We can see that the terms get cancelled successively, so the left term is
= tan-1(n2 + n + 1) - tan-1(1)
\[{\tan ^{ - 1}}\left( {{n^2} + n + 1} \right) - \dfrac{\pi }{4}\]
Hence, proved.
Note: In questions involving summation of tan-1 functions, we should always try to transform it into the form of difference. In this way, the successive terms get cancelled and the summation can be simplified easily.