11.6. Calculate $\lim _{n \rightarrow \infty}\left(\sum_{k=0}^{n} C_{2 n}^{2 k} 61^{k}\right) /\left(\sum_{k=0}^{n-1} C_{2 n}^{2 k+1} 61^{k}\right)$.
Solution. We have $x_{n}=\sum_{k=0}^{n} C_{2 n}^{2 k} 61^{k}=\sum_{k=0}^{n} C_{2 n}^{2 k}(\sqrt{61})^{2 k}$ and $y_{n}=\sum_{k=0}^{n-1} C_{2 n}^{2 k+1} 61^{k}=\frac{1}{\sqrt{61}} \sum_{k=0}^{n-1} C_{2 n}^{2 k+1}(\sqrt{61})^{2 k+1}, \forall n \geq 1$, which implies $x_{n}+y_{n} \sqrt{61}=(1+\sqrt{61})^{2 n}$ and $x_{n}-y_{n} \sqrt{61}=(1-\sqrt{61})^{2 n}, \forall n \geq 1$. We obtain $x_{n}=\frac{(1+\sqrt{61})^{2 n}+(1-\sqrt{61})^{2 n}}{2}$ and $y_{n}=\frac{(1+\sqrt{61})^{2 n}-(1-\sqrt{61})^{2 n}}{2 \sqrt{61}}, \forall n \geq 1$, which implies $\frac{x_{n}}{y_{n}}=\sqrt{61} \cdot \frac{(1+\sqrt{61})^{2 n}+(1-\sqrt{61})^{2 n}}{(1+\sqrt{61})^{2 n}-(1-\sqrt{61})^{2 n}}, \forall n \geq 1$. It follows that

$$
\lim _{n \rightarrow \infty} \frac{\sum_{k=0}^{n} C_{2 n}^{2 k} 61^{k}}{\sum_{k=0}^{n-1} C_{2 n}^{2 k+1} 61^{k}}=\lim _{n \rightarrow \infty} \frac{x_{n}}{y_{n}}=\sqrt{61} \cdot \lim _{n \rightarrow \infty} \frac{(1+\sqrt{61})^{2 n}+(1-\sqrt{61})^{2 n}}{(1+\sqrt{61})^{2 n}-(1-\sqrt{61})^{2 n}}=\sqrt{61} \cdot \lim _{n \rightarrow \infty} \frac{1+\left(\frac{\sqrt{61}-1}{\sqrt{61}+1}\right)^{2 n}}{1-\left(\frac{\sqrt{61}-1}{\sqrt{61}+1}\right)^{2 n}}=\sqrt{61}
$$