[m]S_{n}=\sum_{1}^{n}\frac{12}{(2k-3)(2k+3)}=2\sum_{1}^{n}\frac{1}{2k-3}-\sum_{1}^{n}\frac{1}{2k+3}=[/m]
[m]=2(\frac{1}{1}-\frac{1}{7}+\frac{1}{3}-\frac{1}{9}+\frac{1}{5}-\frac{1}{11}+\frac{1}{7}-\frac{1}{13}-\frac{1}{2k-1}-\frac{1}{2k+1}-\frac{1}{2k+3}[/m]
По определению сумма ряда
S=lim_(n → ∞ )S_(n)=[m]2\cdot(1+\frac{1}{3}+\frac{1}{5}-0-0-0)=\frac{46}{15}[/m]