$\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\dots$ to n terms $=\frac{\mathrm{kn}}{\mathrm{n}+1}$then k=

$$\begin{gathered} \frac{1}{2·4}+\frac{1}{4·6}+\frac{1}{6·8}+.......=\frac{\mathrm{kn}}{\mathrm{n}+1} \\ \Rightarrow 2 \left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.......+\frac{1}{2\mathrm{n}}-\frac{1}{2\mathrm{n}+2}\right]=\frac{\mathrm{kn}}{\mathrm{n}+1} \\ \frac{1}{2} \left[\frac{1}{2}-\frac{1}{2(\mathrm{n}+1)}\right]=\frac{\mathrm{kn}}{\mathrm{n}+1} \\ \frac{1}{4}\left(1-\frac{1}{\mathrm{n}+1}\right)=\frac{\mathrm{kn}}{\mathrm{n}+1} \Rightarrow \frac{1}{4} \left(\frac{\mathrm{n}}{\mathrm{n}+1}\right)=\frac{\mathrm{kn}}{\mathrm{n}+1} \\ \Rightarrow \mathrm{k}=\frac{1}{4} \end{gathered}$$