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Difference between revisions of "2009 AIME I Problems/Problem 7"

m (→‎Solution 2 (Telescoping?))
m (→‎Solution 2 (Telescoping?))
 
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<cmath>a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}</cmath>
 
<cmath>a_{n + 1} - a_n = \log_5{(3n + 5)} - \log_5{(3n + 2)}</cmath>
 
Plug in <math>n = 1, 2, 3, 4</math> to see the first few terms of the sequence: <cmath>\log_5{5},\log_5{8}, \log_5{11}, \log_5{14}.</cmath> We notice that the terms  <math>5, 8, 11, 14</math> are in arithmetic progression. Since <math>a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}</math>, we can easily use induction to show that <math>a_n = \log_5{(3n + 2)}</math>. So now we only need to find the next value of <math>n</math> that makes <math>\log_5{(3n + 2)}</math> an integer. This means that <math>3n + 2</math> must be a power of <math>5</math>. We test <math>25</math>: <cmath>3n + 2 = 25</cmath> <cmath>3n = 23</cmath> This has no integral solutions, so we try <math>125</math>: <cmath>3n + 2 = 125</cmath> <cmath>3n = 123</cmath> <cmath>n = \boxed{041}</cmath>
 
Plug in <math>n = 1, 2, 3, 4</math> to see the first few terms of the sequence: <cmath>\log_5{5},\log_5{8}, \log_5{11}, \log_5{14}.</cmath> We notice that the terms  <math>5, 8, 11, 14</math> are in arithmetic progression. Since <math>a_1 = 1 = \log_5{5} = \log_5{(3(1) + 2)}</math>, we can easily use induction to show that <math>a_n = \log_5{(3n + 2)}</math>. So now we only need to find the next value of <math>n</math> that makes <math>\log_5{(3n + 2)}</math> an integer. This means that <math>3n + 2</math> must be a power of <math>5</math>. We test <math>25</math>: <cmath>3n + 2 = 25</cmath> <cmath>3n = 23</cmath> This has no integral solutions, so we try <math>125</math>: <cmath>3n + 2 = 125</cmath> <cmath>3n = 123</cmath> <cmath>n = \boxed{041}</cmath>
−
== Solution 2 (Telescoping?) ==
+
== Solution 2 (Telescoping) ==
 
We notice that by multiplying the equation from an arbitrary <math>a_n</math> all the way to <math>a_1</math>, we get:
 
We notice that by multiplying the equation from an arbitrary <math>a_n</math> all the way to <math>a_1</math>, we get:
 
<cmath>5^{a_n-a_1}=\dfrac{n+\tfrac23}{1+\tfrac23}</cmath>
 
<cmath>5^{a_n-a_1}=\dfrac{n+\tfrac23}{1+\tfrac23}</cmath>

Latest revision as of 23:26, 8 January 2023

Problem

The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$. Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$.

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

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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