3. Find all natural solutions to the inequality

$$
\frac{4}{5}+\frac{4}{45}+\frac{4}{117}+\cdots+\frac{4}{16 n^{2}-8 n-3}>n-5
$$

In your answer, write the sum of all found solutions.
Solution. We have $16 n^{2}-8 n-3=(4 n-3)(4 n+1), \quad \frac{1}{4 n-3}-\frac{1}{4 n+1}=\frac{4}{16 n^{2}-8 n-3}$,

$\frac{4}{1 \cdot 5}+\frac{4}{5 \cdot 9}+\frac{4}{9 \cdot 13}+\cdots+\frac{4}{(4 n-3)(4 n+1)}>n-5$,

$1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\cdots+\frac{1}{4 n-3}-\frac{1}{4 n+1}>n-5,1-\frac{1}{4 n+1}>n-5,4 n^{2}-23 n-5<0$,

$n \in\left(\frac{23-\sqrt{609}}{8} ; \frac{23-\sqrt{609}}{8}\right) \cap \mathbb{N}, 1+2+3+4+5=15$.

Answer: 15.