Question

$\frac{1}{1.3}+\frac{2}{\mathrm{1.3.5}}+\frac{3}{\mathrm{1.3.5.7}}+\frac{4}{\mathrm{1.3.5.7.9}}+\cdots$

Answer 1

$\begin{array}{c}\frac{1}{1.3}+\frac{2}{\mathrm{1.3.5}}+\frac{3}{\mathrm{1.3.5.7}}+\frac{4}{\mathrm{1.3.5.7.9}}+\cdots \\ T_1=\frac{1}{1.3}=\frac{1}{2}\left[\frac{3-1}{1.3}\right]=\frac{1}{2}\left[1-\frac{1}{3}\right]=\frac{1}{2}-\frac{1}{2.3}\\ T_2=\frac{2}{\mathrm{1.3.5}}=\frac{5-3}{3.5}=\frac{1}{3}-\frac{1}{5}\\ T_3=\frac{3}{\mathrm{3.5.7}}=\frac{1}{2}\left[\frac{7-5}{5.7}\right]=\frac{1}{2}\left[\frac{1}{5}-\frac{1}{7}\right]\\ ⋮\\ T_n=\frac{1}{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]\\ So,\\ S=\frac{1}{2}\sum _{r=1}^n\frac{1}{2r-1}-\frac{1}{2r+1}\\ =\frac{1}{2}\left[\frac{1}{1}-\frac{1}{3}\right]\\ -\frac{1}{2}\left[\frac{1}{1}-\frac{1}{3}\right]\\ \frac{1}{2n-1}-\frac{1}{2n+1}\\ S=\frac{1}{2}\left[1-\frac{1}{2n+1}\right]\\ =\frac{2n}{2\left(2n+1\right)}=\frac{n}{n+1}\\ Therefore,\\ S=\frac{1}{2}\left[1-\frac{1}{\mathrm{1.3.5...}\left(2n+1\right)}\right]\end{array}$