If $$\frac{1}{{2 \times 4}} + \frac{1}{{4 \times 6}} + \frac{1}{{6 \times 8}} + .....\,n{\text{ terms}} = \frac{{kn}}{{n + 1}},$$          then $$k$$ is equal to

Solution :
$$\eqalign{ & \frac{{kn}}{{n + 1}} = \left[ {\frac{1}{{2 \times 4}} + \frac{1}{{4 \times 6}} + \frac{1}{{6 \times 8}} + .....\,n{\text{ terms}}} \right] \cr & = \frac{1}{2}\left[ {\frac{{4 - 2}}{{2 \times 4}} + \frac{{6 - 4}}{{4 \times 6}} + \frac{{8 - 6}}{{6 \times 8}} + ..... + \frac{{2n + 2 - 2n}}{{2n\left( {2n + 2} \right)}}} \right] \cr & = \frac{1}{2}\left[ {\frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{6} + \frac{1}{6} - \frac{1}{8} + ..... + \frac{1}{{2n}} - \frac{1}{{2n + 2}}} \right] \cr & = \frac{1}{2}\left[ {\frac{1}{2} - \frac{1}{{2\left( {n + 1} \right)}}} \right] \cr & = \frac{n}{{4\left( {n + 1} \right)}} \cr & \Rightarrow k = \frac{1}{4} \cr} $$