Recurrence relations for students of the third year of secondary school.

Question

I am not able to solve this problem in order to find a explicit form for the recurrence relation (note: in the original text I can read "a with n" and "a with n-1", but I am not able to format here)

a(0) = 2
3 a(n) = a(n-1) + 6

I have to find the general expression of a(n)

Please note that this problem has been recognized as suitable to a student of the third year of secondary school, so you can't use Laplace Transforms or Differential Equations.

As a reference, the general solution is:

a(n) = (3^(n+1) -1) / (3^n)

Thank you for considering my question.

Answer 1

We can simplify the recursion by adding something to $a$ to eliminate that extra $+6$ term. Since I never just remember what the right thing to do is, we can solve for it.

We define $b(n) = a(n) + k$. Then $a(n) = b(n) - k$ and the recursion is

$$3 (b(n) - k) = (b(n-1)-k) + 6$$ $$ 3b(n) = b(n-1) + (6+2k) $$

So if we pick $k=-3$, then we would have the recursion

$$ 3b(n) = b(n-1) $$

which is pretty easy to solve.

For a more complicated recursion with a multiplier like that, there is a similar trick to eliminate the multiplier. To demonstrate with this example... again, I don't always remember the right thing so I solve for it:

If we set $c(n) = r^n b(n)$, so that $b(n) = r^{-n} c(n)$, then

$$ 3 r^{-n} c(n) = r^{-(n-1)} c(n-1) $$ $$ c(n) = \frac{r}{3} c(n-1) $$

so if we set $r=3$, we would get the recursion

$$ c(n) = c(n-1) $$

which is trivial to solve.

Answer 2

do it step by step: $a_n=\frac{a_{n-1}}{3}+2$ Then $a_{n+1}=\frac{a_{n}}{3}+2$. Now you can insert the expression for $a_n$. $a_{n+1}=\frac{a_{n-1}}{9}+\frac{8}{3}$. You can do it in the same way for $a_{n+2}$ Then set for $a_{n-1}=a_0=2$ and calculate $a_n=a_1,a_{n+1}=a_2,a_{n+2}=a_3$

If you look at the results you might recognize a pattern.

greetings,

calculus

Answer 3

3 a(n) = a(n-1) + 6, a(0) = 2 (warning ==> I need a(1) and I calculate it by hand, the result is: 8/3)

Lowering the indexes of 1, I can write:

3 a(n-1) = a(n-2) + 6

I make the difference between the original and 3 a(n-1) = a(n-2) + 6. I obtain:

3 a(n) - 3 a(n-1) = a(n-1) - a(n-2)

3 a(n) - 4 a(n-1) + a(n-2) = 0

To solve this, I put it in the form:

a(n) = 4/3 a(n-1) - 1/3 a(n-2)

The characteristical equation, in this particolar case, is:

r^2 - 4/3 r + 1/3 = 0

The solutions are real: r = 1/3 and r = 1

The roots are distinct, so the general solution is:

a(n) = x * (r1)^n + y * (r2)^n

Hence:

a(n) = x * (1/3)^n + y * (1)^n

this means:

a(n) = x * (1/3)^n + y

To find a PARTICULAR solution, given the initial conditions, a0 (= 2) and a1 (= 8/3), I have to solve the following system:

ao = x + y a1 = (r1)x + (r2)y

So I can write:

2 = x+y 8/3 = (1/3)*x + y

and then:

y = 8/3 -x/3

This means that:

2 = x + 8/3 -x/3 equal to: 2 = 2x/3 + 8/3 equal to: 6/3 = 2x/3 + 8/3 equal to: 6 = 2x + 8 equal to: 2x = -2

The solutions of the previous system are: x = -1 y = 3

So we have at last:

a(n) = -1 / (3^n) + 3

Factoring:

a(n) = (-1 + 3*3^n) / (3^n)

this means:

a(n) = (3^(n+1) - 1) / (3^n)