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Bulgarian Mathematical Competitions

Algebra Difficulty 5.5 AIME, harder Prove it Bulgaria

Problem:

The sequence $\{a_{n}\}_{n=1}^{\infty}$ 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
$\lim _{n \rightarrow \infty} \frac{\sqrt{a_{n}}+\sqrt{a_{4 n}}+\sqrt{a_{4^{2} n}}+\cdots+\sqrt{a_{4^{10} n}}}{\sqrt{a_{n}}+\sqrt{a_{2 n}}+\sqrt{a_{2^{2} n}}+\cdots+\sqrt{a_{2^{10} n}}}$

Solution

Solution:

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

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