# Sequence and series

Algebra

### Sequence and series

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
Guest

### Re: Sequence and series

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

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

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

So $$d_0= 1$$, $$d_1= 1/5$$, $$d_2= 1/25$$, etc.

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

So it is obvious that $$d_n= \frac{1}{5^n}$$.

Now we switch around and define $$s_n= u_n+ v_n$$ rather than subtracting.

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

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

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

Since $$s_n= v_n+ u_n$$ and $$d_n= v_n- u_n$$, adding the two $$s_n+ d_n= 2v_n$$.
And subtracting, $$s_n- d_n= 2u_n$$.

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

HallsofIvy

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