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 1


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.

  • $\begingroup$ 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 $\endgroup$
    – Dan
    Dec 31, 2020 at 10:09
  • $\begingroup$ @Dan I have accordingly corrected. $\endgroup$
    – Allawonder
    Dec 31, 2020 at 10:12
  • $\begingroup$ So i calculated un = 1.4 and vn = 1.6 ? $\endgroup$
    – Dan
    Dec 31, 2020 at 10:16
  • $\begingroup$ @Dan If your computations were correct then those are constant sequences, which you should be able to solve to finish off the problem $\endgroup$
    – Allawonder
    Dec 31, 2020 at 10:23
  • $\begingroup$ im sorry but i didnt get it, how can i find their expression please $\endgroup$
    – Dan
    Dec 31, 2020 at 10:35

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