Question

Sum the series to $n$ terms
$\frac{1}{\mathrm{2.4.6}}+\frac{1}{\mathrm{4.6.8}}+\frac{1}{\mathrm{6.8.10}}+...$

Answer 1

$\frac{1}{\mathrm{2.4.6}}+\frac{1}{\mathrm{4.6.8}}+\frac{1}{\mathrm{6.8.10}}+....$

$v_0=\frac{1}{\mathrm{2.4.6}}=\frac{1}{4}\{\frac{1}{2.4}-\frac{1}{4.6}\}$

$V_1=\frac{1}{\mathrm{4.6.8}}=\frac{1}{4}\{\frac{1}{4.6}-\frac{1}{6.8}\}$

$\frac{1}{\mathrm{6.8.10}}=\frac{1}{4}\{\frac{1}{6.8}-\frac{1}{8.10}\}$

$V_n=\frac{1}{2n(2n+2)(2n+4)}=\frac{1}{4}\{\frac{1}{\left(2n\right)(2n+2)}-\frac{1}{(2n+2)(2n+4)}\}$

Adding these two all above eq${}^n$, we get

$=\frac{1}{4}[\frac{1}{2.4}-\frac{1}{(2n+2)(2n+4)}]$

$S_n=\frac{1}{(r-1)d}(V_0-V_n)$

$S_n=\frac{1}{2.2}\{\frac{1}{2.4}-\frac{1}{(2n+2)(2n+4)}\}$