$\sum_{n=1}^{\infty}=\frac{1}{16n^2-8n-3} \Rightarrow s_n=\sum_{k=1}^{n}\frac{1}{16k^2-8k-3}$
$\frac{1}{16k^2-8k-3}=\frac{1}{(4k-3)(4k+1)}=\frac{1}{4}\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)$
$\Rightarrow \sum_{k=1}^{n}\frac{1}{4}\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)=\frac{1}{4}\left(1-\frac{1}{4n+1}\right) \ \ \ (*)$
$\Rightarrow S=\lim_{n \to \infty} \frac{1}{4}\left(1-\frac{1}{4n+1}\right)=\frac{1}{4}$
The question is about the (*) line. Why does the first member equal the second one?
Thank you.