We deposit $2000$ dollars in an investment fund, a year later we have $2500$ dollars between investment and earnings. Earnings from the year $(n-1)$ to the year $(n)$ triple the earnings from the year $(n-2)$ to the year $(n-1)$
Write the recurrence relation and an initial condition that defines the amount of money over time. Then solve it.
Take into account the following variables and syntax to represent the terms of the sequence
$n$: year
$a_n$: amount of money you have in year $n$
I don't even understand this problem, any help would be much appreciated!
Recurrence Relation
$$a_n – a_{n-1} = 3(a_{n-1} – a_{n-2})$$ $$\tag{1}a_n = 4(a_{n-1}) – 3(a_{n-2})$$
Initial conditions
$$a_0=2000$$ $$a_1=2500$$
Solving
For the first-order recurrence relation, the solution was of the form: $$S_n=t^n$$ Then for our first attempt at finding a solution of the second-order recurrence relation $[1]$, we will search for a solution of the form: $$V_n = t^n$$
If $V_n = t^n$ is to solve $[1]$, we must have $$V_n = 4V_{n−1} − 3V_{n−2}$$ or $$t^n = 4t^{n−1} − 3t^{n−2}$$ or $$t^n - 4t^{n−1} + 3t^{n−2}=0$$ Dividing by $t^{n−2}$, we obtain the equivalent equation $$t^2 − 4t + 3 = 0 ,[2]$$ Solving $[2]$, we find the solutions $$t = 1, t = 3$$ At this point, we have two solutions S and T of $[1]$, given by $$S_n = 1^n=1 , T_n = 3^n$$
We can verify (see Theorem 7.2.11 - Discrete Mathematics 8th edition by Johnsonbaugh) that if S and T are solutions of $[1]$, then bS+ dT, where b and d are any numbers whatever, is also a solution of $[1]$. In our case, if we define the sequence U by the equation $$U_n = bS_n + dT_n = b + d*3^n$$ , U is a solution of $[1]$. To satisfy the initial conditions, we must have $$2000 = U_0 = b + d*3^0 = b + d$$ $$2500 = U_1 = b + d*3^1 = b + 3d$$ Solving these equations for b and d, we obtain $b = 1750$ and $d = 250$. Therefore, the sequence $U$ defined by $$U_n = 1750 + 250 ·3^n$$ satisfies the recurrence relation $[1]$ and the initial conditions. We conclude that $$a_n = U_n = 1750 + 250 ·3^n$$ for $n = 0, 1,...$