Question

Find the sum of n terms of the series $\frac{1}{1.3}+\frac{2}{\mathrm{1.3.5}}+\frac{3}{\mathrm{1.3.5.7}}+\frac{4}{\mathrm{1.3.5.7.9}}+...$

Answer 1

$t_n=\frac{n}{1\cdot 3\cdot 5\cdot \cdot \cdot \cdot \cdot \cdot (2n-1)(2n+1)}$

$=\frac{A(n+1)+B}{1\cdot 3\cdot 5\cdot \cdot \cdot \cdot (2n+1)}-\frac{An+B}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)}$ (say)

$n=An+A+B-(An+B)(2n+1)$

By equating co efficients

$A=0,B=-\frac{1}{2}$

$\therefore t_n=\frac{1}{2}\frac{1}{1\cdot 3\cdot 5\cdot \cdot \cdot \cdot (2n-1)}-\frac{1}{2}\frac{1}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n+1)}$

$t_{n-1}=\frac{1}{2}\frac{1}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n-3)}-\frac{1}{2}\frac{1}{1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)}$

$t_2=\frac{1}{2}\frac{1}{1\cdot 3}-\frac{1}{2}\frac{1}{1\cdot 3\cdot 5}$

$t_1=\frac{1}{2}\cdot \frac{1}{1}-\frac{1}{2}\frac{1}{1\cdot 3}$

$S_n=\frac{1}{2}-\frac{1}{2}\cdot \frac{1}{1\cdot 3\cdot 5\cdot \cdot \cdot (2+1)}$