We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
162
7
avatar

I have found the answer (41) to this problem by virtue of using logarithms, however I’m in need of a solution/proof using telescoping/sequences ! I would really appreciate it, and I need the solution quickly. This is the problem

 

 Jan 13, 2019
edited by Guest  Jan 13, 2019
 #3
avatar+533 
0

 

i suppose you could try plugging in 1, 2, 3, 4 and seeing if there is a pattern

 

EDIT: nevermind i read want you wanted

 Jan 13, 2019
edited by asdf335  Jan 13, 2019
 #5
avatar+7531 
+1

\(5^{a_{n+1}-a_n} - 1 = \dfrac{1}{n+\frac{2}{3}}, n\geq 1\)

\(k > 1, k\in \mathbb Z^+\) and \(a_k\) is an integer.

Let us do something to the recursive definition of the sequence.

\(\dfrac{5^{a_{n+1}}-5^{a_n}}{5^{a_n}} = \dfrac{1}{n+\frac{2}{3}}\\ 5^{a_{n+1}} = \left(\dfrac{3}{3n+2}+1\right) \cdot 5^{a_n}\\ 5^{a_{n+1}} = \left(\dfrac{3n+5}{3n+2}\right) \cdot 5^{a_n}\\ a_{n+1} = a_n + \log_5\left(\dfrac{3(n+1)+2}{3n+2}\right)\)

You see somehow it is telescoping :P

Now list some values of an for small n.

\(a_1 = 1\\ a_2 = 1 + \log_5\left(\dfrac{8}{5}\right) = \log_5(8)\\ a_3 = 1 + \log_5\left(\dfrac{8}{5}\right) + \log_5\left(\dfrac{11}{8}\right)=1+\log_5\left(\dfrac{11}{5}\right)=\log_5(11)\\\)

You observe that: \(a_n = \log_5\left(3n+2\right)\)

Now we got the explicit formula for an. We are halfway done :) (Nice :D)

If ak is an integer, then \({3k+2}\) must be a power of 5.

Now the question becomes a number theory question: What is the least value of k such that k > 1 and 3k + 2 is a power of 5?

Let \(5^x = 3k+2\).

\(5^x \equiv 2 \pmod 3\)

\(2^x \equiv 2 \pmod 3\)

This congruence holds true when x is an odd integer.

So the least value of k can be obtained by the equation \(5^3 = 3k+2\). (Why not 51?)

\(3k + 2 = 125\\ k = \dfrac{123}{3} = 41\)

Voilà!

 Jan 14, 2019
edited by MaxWong  Jan 14, 2019
edited by MaxWong  Jan 14, 2019
edited by MaxWong  Jan 14, 2019
 #6
avatar+101741 
+3

Nice answer Max :)

Melody  Jan 14, 2019
 #7
avatar+7531 
+1

Thank you ;)

MaxWong  Jan 14, 2019

19 Online Users

avatar
avatar
avatar
avatar