Answer to Question #156260 in Calculus for Terry

Question

We consider the sequence of real numbers (Un) defined on N by Uo = -1, U1 = 1/2 and for every n E N, U(n+2) = U(n+1) - 1/4 Un. Where N represents the set of natural numbers
(i) Calculate Vo.
(ii) Show that (Vn) is a geometric sequence and precise its common ratio.
(iii) Express Vn in terms of n and calculate its limit.

Answer 1

U(n+2) = U(n+1) - 1/4 Un

Vn = U(n+1) - (1/2)Un.

(i)

$$V(0)=U(0+1)-(1/2)U(0)$$

$$=1/2-(1/2)(-1)=1$$

(ii)

$U(0)=-1,$

$U(1)=1/2,$

$U(2)=1/2+1/4=3/4$

$U(3)=3/4-1/8=5/8$

$U(4)=5/8-1/8=7/16$

$U(5=7/16-5/32=9/32$

$...$

$U(k)=(2k-1)/2^k$

Prove by mathematical induction

Let the statement $P(n)$ given as

$P(n):U(n)=(2n-1)/2^n$

Then

$P(0):U(0)=(2(0)-1)/2^0=-1$ is true.

$P(1): U(1)=(2(1)-1)/2^1=1/2$ is true.

Assume that $P(n)$ is true for some natural number $k$ , i.e.,

$P(k):U(k)=(2k-1)/2^k$

$P(k+1):U(k+1)=(2(k+1)-1)/2^{k+1}$

Now, to prove that $P(k+2)$ is true, we have

$$P(k+2): U(k+2)=U(k+1)-(1/4)U(k)=$$

$$=\dfrac{2(k+1)-1}{2^{k+1}}-\dfrac{2k-1}{(4)(2^{k})}=\dfrac{4k+4-2-2k+1}{2^{k+2}}$$

Thus $P(k+2)$ is true, whenever $P(k)$ and $P(k+1)$ is true. Hence, by the Principle of Mathematical Induction $P(n)$ is true for all natural numbers $n$

$$U(n)=(2n-1)/2^n, n=0,1,2,...$$

Then

$$V(n)=U(n+1)-(1/2)U(n)$$

$$=\dfrac{2(n+1)-1}{2^{n+1}}-\dfrac{2n-1}{2^{n+1}}=\dfrac{1}{2^n}, n=0,1,2,...$$

Then $(V(n))$ is a geometric sequence and its common ratio is $1/2$

(iii)

$$V(n)=\dfrac{1}{2^n}, n=0,1,2,...$$

$$\lim\limits_{n\to \infin}V(n)=\lim\limits_{n\to \infin}\dfrac{1}{2^n}=0$$