Giải thích các bước giải:
$u_n=\sum^n_{k=1}\dfrac{1}{9k^2-1}$
$\to u_n=\sum^n_{k=1}\dfrac{1}{(3k-1)(3k+1)}$
$\to u_n=\dfrac 12\sum^n_{k=1}\dfrac{2}{(3k-1)(3k+1)}$
$\to u_n=\dfrac 12\sum^n_{k=1}\dfrac{(3k+1)-(3k-1)}{(3k-1)(3k+1)}$
$\to u_n=\dfrac 12\sum^n_{k=1}\dfrac{1}{3k-1}-\dfrac{1}{3k+1}$
$\to u_n=\dfrac 12(\dfrac{1}{3.1-1}-\dfrac{1}{3n+1})$
$\to u_n=\dfrac 12(\dfrac{1}{2}-\dfrac{1}{3n+1})$
$\to u_n=\dfrac{3n-1}{2(3n+1)}$