Solving a sequence

Question

Given a sequence $$x_0=5$$ $$x_{n+1}=x_n*1.03+1$$ how does one find generalized formulae for:

1. $x_n$
2. $\sum_{n=a}^{b}x_n$ for a given pair of $a, b$ where $a{\leq}b$ and $a>0$

Interested in both the closed formulae and (a wonderful cherry on top!) a description on the methodology of solving these kinds of problems.

I can obviously do this recursively, but it feels like this is a common enough problem that has been solved many times over. I'm just unable to find the solution by googling (most likely due to lack of math education - what terms should I be looking for besides the "sequence" and "solve"?)

Answer 1

Your first problem may be solved by induction on $n \in \mathbb{N}$. The general formula is: $x_{n} = 5*(1.03)^{n} + \sum_{j=1}^{n}{(1.03)^{j-1}}$. Then you have to verify that: $(i)$ The formula holds for $n = 1$. $(ii)$ If the formula holds for $n \in \mathbb{N}$ then it holds for $n+1$. Which is not hard to verify. Your second problem is a sum of a geometric progression. Can you conclude from here?

Answer 2

$$x_0=5$$ $$x_{n+1}=1.03x_n+1\tag{1}$$ plug $n+1\to n$ $$x_{n+2}=1.03x_{n+1}+1\tag{2}$$ Subtract $(2)-(1)$ $$x_{n+2}-x_{n+1}=1.03x_{n+1}-1.03x_n$$ rearrange the equation $$x_{n+2}-2.03x_{n+1}+1.03x_n=0$$ solve the characteristic equation $$t^2-2.03t+1.03=0\to t_1=1;\;t_2=1.03$$ general solution is $$x_n=a\cdot 1^n+b\cdot 1.03^n$$

compute $x_1=1.03x_0+1=6.15$

Plug the initial values to get $a,b$

$$a+1.03 b=6.15,a+ b=5\to a=-33.33;\;b=38.33$$ the solution is $$x_n=38.33\cdot 1.03^n-33.33$$

---

$$\sum _{n=p}^q x_n=\frac{5}{9} \left(60 p-23\cdot 100^{1-p} \cdot103^p+23\cdot 100^{-q} \cdot 103^{q+1}-60 q-60\right)$$