Sequence and series (6)

Given that $5^{a_{n+1}-a_n} = \dfrac 3{3n+2} + 1$ , we have:

$\begin{aligned} \frac {5^{a_{n+1}}}{5^{a_n}} & = \frac 3{3n+2} + 1 & \small \blue{\text{Let }b_k = 5^{a_k}} \\ b_{n+1} & = \left(\frac 3{3n+2} + 1 \right)b_n \\ \implies b_2 & = \left(\frac 3{3+2} + 1 \right)5^1 = 3 + 5 = 8 \\ b_3 & = \left(\frac 3{8} + 1 \right)8 = 3 + 8 = 11 \\ b_4 & = \left(\frac 3{11} + 1 \right)11 = 3 + 11 = 14 \end{aligned}$

Therefore, it would appears that $b_n = 3(n-1) + 5 = 3n + 2$ , which can be easily proven by induction , because $\\ b_{n+1} = \left(\dfrac 3{3n+2} + 1 \right)b_n = \left(\dfrac 3{3n+2} + 1 \right)(3n+2) = 3 + 3n+2 = 3(n+1) + 2$ .

Since $b_m = 5^{a_m}$ , for $a_m$ to be an integer, $b_m = 3m+2$ must be a power of $5$ . $3m+2= 5^1$ is when $m=1$ . For higher power of $5$ , $3m+2 \equiv 0 \text{ (mod 25)} \implies m \equiv \boxed{41}$ .