# 2013-17 Repeated numbers

A real sequence $$x_1, x_2, x_3, \cdots$$ satisfies the relation $$x_{n+2} = x_{n+1} + x_n$$ for $$n = 1, 2, 3, \cdots$$. If a number $$r$$ satisfies $$x_i = x_j = r$$ for some $$i$$ and $$j$$ $$(i \neq j)$$, we say that $$r$$ is a repeated number in this sequence. Prove that there can be more than $$2013$$ repeated numbers in such a sequence, but it is impossible to have infinitely many repeated numbers.

GD Star Rating

# 2013-09 Inequality for a sequence

Let $$N > 1000$$ be an integer. Define a sequence $$A_n$$ by
$A_0 = 1, \, A_1 = 0, \, A_{2k+1} = \frac{2k}{2k+1} A_{2k} + \frac{1}{2k+1} A_{2k-1}, \, A_{2k} = \frac{2k-1}{2k} \frac{A_{2k-1}}{N} + \frac{1}{2k} A_{2k-2}.$
Show that the following inequality holds for any integer $$k$$ with $$1 \leq k \leq (1/2) N^{1/3}$$.
$A_{2k-2} \leq \frac{1}{\sqrt{(2k-2)!}}.$

GD Star Rating

# 2012-19 A limit of a sequence involving a square root

Let $$a_0=3$$ and $$a_{n}=a_{n-1}+\sqrt{a_{n-1}^2+3}$$ for all $$n\ge 1$$. Determine $\lim_{n\to\infty}\frac{a_n}{2^n}.$

GD Star Rating

# 2011-24 (n-k) choose k

Evaluate the sum $\sum_{k=0}^{[n/2]} (-4)^{n-k} \binom{n-k}{k} ,$ where [x] denotes the greatest integer less than or equal to x.

GD Star Rating

# 2011-6 Equal sums

Let $$a_1\le a_2\le \cdots \le a_k$$ and $$b_1\le b_2\le \cdots \le b_l$$ be sequences of positive integers at most M. Prove that if $\sum_{i=1}^{k} a_i^n = \sum_{j=1}^l b_j^n$ for all $$1\le n\le M$$, then $$k=l$$ and $$a_i=b_i$$ for all $$1\le i\le k$$.

GD Star Rating

# 2009-4 Initial values

Let $$a_0=a$$ and $$a_{n+1}=a_n (a_n^2-3)$$. Find all real values $$a$$ such that the sequence $$\{a_n\}$$ converges.

GD Star Rating

# 2009-2 Sequence of Log

Let $$a_1<\cdots$$ be a sequence of positive integers such that $$\log a_1, \log a_2,\log a_3,\cdots$$ are linearly independent over the rational field $$\mathbb Q$$. Prove that $$\lim_{k\to \infty} a_k/k=\infty$$.

GD Star Rating
Let $$a_1=\sqrt{1+2}$$,
$$a_2=\sqrt{1+2\sqrt{1+3}}$$,
$$a_3=\sqrt{1+2\sqrt{1+3\sqrt{1+4}}}$$, …,
$$a_n=\sqrt{1+2\sqrt{1+3\sqrt {\cdots \sqrt{\sqrt{\sqrt{\cdots\sqrt{1+n\sqrt{1+(n+1)}}}}}}}}$$, … .
Prove that $$\displaystyle\lim_{n\to \infty} \frac{a_{n+1}-a_{n}}{a_n-a_{n-1}}=\frac12$$.