# Geometric sequence and series problem

Hello i need help please, $$\begin{cases} u_0&= 1\\ u_{n+1}&=\frac{3u_n + 2v_n}{5} \end{cases} \qquad \begin{cases} v_0 = 2\\ v_{n+1} = \frac{2u_n+3v_n}{5} \end{cases}$$

1. Calculate $$u_1$$, $$u_2$$, $$v_1$$, and $$v_2$$.

2. We consider the sequence $$(d_n)$$ defined for any natural number $$n$$ by $$d_n = v_n-u_n$$.

a. Show that the sequence $$(d_n)$$ is a geometric sequence of which we will give its common ratio and its first term.

b. Deduce the expression of $$d_n$$ depending on $$n$$.

3. We consider the sequence (s_n) defined for any natural number $$n$$ by $$s_n = u_n+v_n$$.

a. Calculate $$s_0$$, $$s_1$$, and $$s_2$$. What can we guess?

b. Show that, for all $$n \in \mathbb{N}$$, $$s_{n+1} = s_n$$. What can we deduce? Deduce an expression of $$u_n$$ and $$v_n$$ depending on $$n$$.

5. Determine depending on $$n \in \mathbb{N}$$,

a. $$T_n = u_0 + u_1 + \dots + u_n$$.

b. $$W_n = v_0 + v_1 + \dots + v_n$$.

Actually i answered the first question and the second, and i found the common ratio is $$1/5$$ and the first term 1 so $$d_n = 1 \cdot 1/5^n$$ For the 3. I found $$3$$ And i cant resolve the 4 and 5 can you help me

## 1 Answer

Hints. Since you have successfully found $$d_n=v_n-u_n$$ and $$s_n=v_n+u_n$$ as functions of $$n,$$ then you can find expressions for $$u_n$$ and $$v_n$$ simply by subtraction and addition of the above equations, so that you have that $$2v_n=s_n+d_n$$ and $$2u_n=s_n-d_n.$$

From here it should be easy to continue.

• But dn is dn= vn- un and can just olease show me the first step because i don’t really understand how to do it
– Dan
Dec 31, 2020 at 10:09
• @Dan I have accordingly corrected. Dec 31, 2020 at 10:12
• So i calculated un = 1.4 and vn = 1.6 ?
– Dan
Dec 31, 2020 at 10:16
• @Dan If your computations were correct then those are constant sequences, which you should be able to solve to finish off the problem Dec 31, 2020 at 10:23
• im sorry but i didnt get it, how can i find their expression please
– Dan
Dec 31, 2020 at 10:35