+0

# HELP! Sequence Problem!

0
81
7

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
+429
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
+7220
+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
+97575
+3

Nice answer Max :)

Melody  Jan 14, 2019
#7
+7220
+1

Thank you ;)

MaxWong  Jan 14, 2019