Sequence and series

[Guest]

Hello i need help please,
u0= 1
un+1=( 3un + 2vn)/5
v0 = 2
vn+1 = (2un+3vn)/5
1.Calculate u1,u2,v1,v2
2.We consider the sequence (dn) defined for any natural number n by dn = vn-un
a. Show that the sequence (dn) is a geometric sequence of which we will give its common ratio and its first term.
b. Deduce the expression of dn depending on n
3. We consider the sequence (sn) Defined for any natural number n by sn = un+ vn
a. Calculate s0, s1and s2 . What can we guess?
b. Show that, for all nEN, sn+1= sn. What can we deduce?
4. Deduce an expression of un and vn depending on n.
5.determine depending on n E N
a. Tn= u0 + u1 + ... + un
b. Wn= v0 + v1 + ... + vn

actually i answered the first question and the second and i found the common ratio is 1/5 and the first term 1 so dn= 1*1/5^n
For the 3. I found 3
And i cant resolve the 4 and 5 can you help me

[HallsofIvy]

Alright, we are given that [tex]u_0= 1[/tex] and [tex]v_0= 2[/tex]. We are give that [tex]u_{n+ 1}= (3u_n+ 2v_n)/5[/tex] and that [tex]v_{n+ 1}= (2u_n+ 3v_n)/5[/tex].

Yes, calculating [tex]u_1[/tex], [tex]v_1[/tex], [tex]u_2[/tex], and [tex]v_2[/tex] is straight forward arithmetic.
[tex]u_1= (3u_0+ 2v_0)/5= (3+ 4)/5= 7/5[/tex]
[tex]v_1= (2u_0+ 3v_0)/5= (2+ 6)/5= 8/5[/tex]
[tex]u_2= (3u_1+ 2v_1)/5= (21/5+15/5)/5= 36/25[/tex]
[tex]v_2= (2u_1+ 3v_1)/5= (14/5+ 24/5)/5= 38/25[/tex]

Now what about [tex]d_n= v_n- u_n[/tex]?
Well, [tex]d_0= v_0- u_0= 2- 1= 1[/tex] and [tex]d_{n+ 1}= v_{n+1}- u_{n+1}= (2u_n+ 3v_n)/5- (3u_n+ 2v_n)/5= (v_n- u_n)/5= d_n/5[/tex].

So [tex]d_0= 1[/tex], [tex]d_1= 1/5[/tex], [tex]d_2= 1/25[/tex], etc.

Yes, [tex]d_n[/tex] is a geometric series with first term 1 and common ratio 1/5.

So it is obvious that [tex]d_n= \frac{1}{5^n}[/tex].

Now we switch around and define [tex]s_n= u_n+ v_n[/tex] rather than subtracting.

We have [tex]s_0= u_0+ v_0= 1+ 2= 3[/tex], [tex]s_1= u_1+ v_1= 7/5+ 8/5= 15/5= 3[/tex], and [tex]s_2= u_2+ v_2= 36/25+ 38/25= 75/25= 3[/tex].

In fact, [tex]s_{n+1}= u_{n+1}+ v_{n+1}= (3u_n+ 2v_n)/5+ (2u_n+ 3v_n)/5= u_n+ v_n= s_n[/tex].

Well, that's pretty darn easy! The first term is 3 and it doesn't change. [tex]s_n= 3[/tex] for all n.

Since [tex]s_n= v_n+ u_n[/tex] and [tex]d_n= v_n- u_n[/tex], adding the two [tex]s_n+ d_n= 2v_n[/tex].
And subtracting, [tex]s_n- d_n= 2u_n[/tex].

So [tex]u_n= \frac{s_n- d_n}{2}[/tex] and [tex]v_n= \frac{s_n+ d_n}{2}[/tex]. Use your formulas for [tex]d_n= \frac{1}{5^n}[/tex] and [tex]s_n= 3[/tex] to complete that.