# A recurrence Relations problem that I found in a manga

So this question is from the manga series, “Assassination Classroom”. The Japanese Roughly above given on the cyborg translates to: “Find the general solution for a[n] as defined in following recurrence-relation sequence.”

The recurrence relation sequence in this scan looks something is shown at the bottom of that cyborg

Here’s a drawing of mine trying to depict said recurrence relation clearly.

I’m not sure if something like that would count as a recurrence relation or if the vertical “=“ signs are actually quotations (“ “)

Anyway, a character “solves” the question and has written the solution (or at least, a trick) on a grenade.: a[n+1]/3[n+1] and a[n]/3[n]

So, can somebody help me interpret what the recurrence relation in the question is and what the steps are to finding the solution?

Thank you.

• I don't understand the kind of information that can be exploited in your drawing neither the meaning that is attributed to $3[n]$ or $3[n+1]$ Jun 10 at 17:00
• I appreciate the attempt, but I think there’s not enough information here to determine if this is real math or just a bunch of random symbols.
– Eric
Jun 10 at 17:17

Sometimes japanese is written vertically top to bottom, right to left. So the recurrence in the manga is \begin{align} a_{1} &= 5, \\ a_{n+1} &= 2a_{n} + 3. \end{align} I don't see any relation between the symbols on the grenade and the solution the recurrence.
This can be solved by subtituting $$a_{n}-2a_{n-1}$$ into $$3,$$ which yields $$a_{n+1} = 3a_n - 2a_{n-1}.$$ This is a homogeneous linear recurrence, here is one way to solve it. The final solution of this recurrence is $$a_{n} = 2^{n+2}-3.$$
• Another way to solve this is to add 3 to both sides of the initial recurrence to get $a_{n+1}+3=2(a_n+3)$. So $a_n+3$ doubles on each iteration, and since $a_1+3=8=2^3$ we have $a_n+3=2^{n+2}$. Jun 10 at 19:03