Problem 11.1. The sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ is defined by $a_{1}=0$ and $a_{n+1}=$ $a_{n}+4 n+3, n \geq 1$.

a) Express $a_{n}$ as a function of $n$.

b) Find the limit

$$
\lim _{n \rightarrow \infty} \frac{\sqrt{a_{n}}+\sqrt{a_{4 n}}+\sqrt{a_{4^{2} n}}+\cdots+\sqrt{a_{4}{ }^{10} n}}{\sqrt{a_{n}}+\sqrt{a_{2 n}}+\sqrt{a_{2^{2} n}}+\cdots+\sqrt{a_{2^{10} n}}}
$$

Emil Kolev
11.1. a) Using the recurrence relation we easily get

$$
\begin{aligned}
a_{k} & =a_{k-1}+4(k-1)+3=a_{k-2}+4(k-2)+4(k-1)+2 \cdot 3=\cdots \\
& =a_{1}+4(1+2+\cdots+k-1)+(k-1) \cdot 3=2 k(k-1)+3(k-1) \\
& =(2 k+3)(k-1)
\end{aligned}
$$

b) We have $\lim _{n \rightarrow \infty} \frac{\sqrt{a_{k n}}}{n}=\lim _{n \rightarrow \infty} \sqrt{\left(2 k+\frac{3}{n}\right)\left(k-\frac{1}{n}\right)}=\sqrt{2} k$. Therefore the required limit is equal to

$$
\frac{1+4+4^{2}+\cdots+4^{10}}{1+2+2^{2}+\cdots+2^{10}}=\frac{4^{11}-1}{3\left(2^{11}-1\right)}=\frac{2^{11}+1}{3}=683
$$