4. Let

$$
a_{1}=\frac{2}{3} \text { and } a_{n+1}=\frac{a_{n}}{4}+\sqrt{\frac{24 a_{n}+9}{253}}-\frac{9}{48}
$$

for every natural number $n$. Calculate the sum $a_{1}+a_{2}+a_{3}+\ldots$.
Solution. Let $b_{n}=\sqrt{24 a_{n}+9}$. Then we have $a_{n}=\frac{b_{n}^{2}-9}{24}$, so the initial recurrence relation takes the form

$$
\frac{b_{n+1}^{2}-9}{24}=\frac{b_{n}^{2}-9}{24}+\frac{b_{n}}{16}-\frac{9}{48} \text {, i.e. }\left(2 b_{n+1}\right)^{2}=\left(b_{n}+3\right)^{2}
$$

Since $b_{n} \geq 0$, for every $n \in \mathbb{N}$, it follows that $2 b_{n+1}=b_{n}+3$. The last equality can be written as

$$
2\left(b_{n+1}-3\right)=b_{n}-3
$$

We set $c_{n}=b_{n}-3$ and get

$$
c_{n+1}=\frac{1}{2} c_{n} \text { and } c_{1}=b_{1}-3=\sqrt{24 a_{1}+9}-3=2
$$

Thus,

$$
c_{n}=\frac{1}{2^{n-2}}, \text { so } b_{n}=\frac{1}{2^{n-2}}+3 \text { and } a_{n}=\frac{1}{24} \cdot \frac{1}{2^{2 n-4}}+\frac{1}{2^{n}}
$$

Finally,

$$
a_{1}+a_{2}+a_{3}+\ldots=\frac{1}{24}\left(4+1+\frac{1}{4}+\ldots\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots\right)=\frac{1}{24} \cdot \frac{4}{1-\frac{1}{4}}+\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{11}{9}
$$