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

Guest Jan 13, 2019

edited by
Guest
Jan 13, 2019

#5**+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 a_{n} 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 a_{n}. We are halfway done :) (Nice :D)

If a_{k} 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 5^{1}?)

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

Voilà!

MaxWong Jan 14, 2019