$\text{ The sum of the first }10\text{-terms of the series }\frac{5}{\mathrm{1.2.3}}+\frac{7}{\mathrm{2.3.9}}+\frac{9}{\mathrm{3.4.27}}+\dots \text{ is }$

${\mathrm{T}}_{\mathrm{n}}=\frac{n^{th}\text{ term of the series }}{n(n+1)\cdot 3^n}=\frac{2n+3}{n(n+1)3^n}$

$\begin{array}{l}\mathrm{Let}\frac{2n+3}{n(n+1)3^n}=\left(\frac{3}{n}-\frac{1}{n+1}\right)\frac{1}{3^n}\\ =\frac{1}{n{3}^{n-1}}-\frac{1}{3^n(n+1)}\\ S_n=\sum T_n=\sum \left[\frac{1}{n{3}^{n-1}}-\frac{1}{3^n(n+1)}\right] =\left(1-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{{3.3}^2}\right)+..+\left[\frac{1}{n\cdot 3^n-1}-\frac{1}{(n+1)3^n}\right]\\ =\left[1-\frac{1}{(n+1)3^n}\right]\\ S_{10}=1-\frac{1}{{11.3}^{10}}\end{array}$