AIME 2009 I · 第 7 题

Solution

The best way to solve this problem is to get the iterated part out of the exponent:

$$5^{(a_{n + 1} - a_n)} = \frac {1}{n + \frac {2}{3}} + 1$$ $$5^{(a_{n + 1} - a_n)} = \frac {n + \frac {5}{3}}{n + \frac {2}{3}}$$ $$5^{(a_{n + 1} - a_n)} = \frac {3n + 5}{3n + 2}$$ $$a_{n + 1} - a_n = \log_5{\left(\frac {3n + 5}{3n + 2}\right)}$$ $$a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}$$ Plug in $n = 1, 2, 3, 4$ to see the first few terms of the sequence:

$$\log_5{5},\log_5{8}, \log_5{11}, \log_5{14}.$$ We notice that the terms $5, 8, 11, 14$ are in arithmetic progression. Since $a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}$, we can easily use induction to show that $a_n = \log_5{(3n + 2)}$. So now we only need to find the next value of $n$ that makes $\log_5{(3n + 2)}$ an integer. This means that $3n + 2$ must be a power of $5$. We test $25$:

$$3n + 2 = 25$$ $$3n = 23$$ This has no integral solutions, so we try $125$:

$$3n + 2 = 125$$ $$3n = 123$$ $$n = \boxed{041}$$

Solution 2 (Telescoping)

We notice that by multiplying the equation from an arbitrary $a_n$ all the way to $a_1$, we get:

$$5^{a_n-a_1}=\dfrac{n+\tfrac23}{1+\tfrac23}$$ This simplifies to

$$5^{a_n}=3n+2.$$ We can now test powers of $5$.

$5$ - that gives us $n=1$, which is useless.

$25$ - that gives a non-integer $n$.

$125$ - that gives $n=\boxed{41}$.

-integralarefun

Solution 3 (I did too many FE's)

The given recursion is equivalent to $\frac{5^{a_{n+1}}}{3(n+1)+2}=\frac{5^{a_n}}{3n+2}$. By a simple inductive argument, the value $\frac{5^{a_k}}{3k+2}$ is constant for all naturals $k$. In particular, when $k=1$, the expression is equal to $1$, so $\frac{5^{a_k}}{3k+2}=1$ for all naturals $k$. Therefore, $a_k=\log_5(3k+2)$, and testing powers of $5$ yields $k=\boxed{41}$ as the answer, which is when $a_{41}=\log_5(3\cdot41+2)=\log_5(125)=3$.

~ eevee9406