Answer to Question #156266 in Calculus for Keleko

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 reprents the set of natural numbers. Vn = U(n+1) - (1/2)Un.
We define the sequence (Wn) by for every n E N, (Wn) = Un / Vn.
(i) Calculate Wo and show that Wn is an arithmetic sequence and precise its common difference.
(ii) Express Wn in terms of n and calculate its limit.

Answer 1

Answer:

Step 1

U₀= -1

U₁= $1 \over 2$

U_n+2= U_n+1 - $1 \over 4$U_n

U₂=U₁- $1 \over 4$U₀ = $1 \over 2$- $1 \over 4$(-1) =  $1 \over 2$+ $1 \over 4$ = $3 \over 4$

U₃=  U₂ - $1 \over 4$U₁= $3 \over 4$ - $1 \over 4$ ( $1 \over 2$) =  $3 \over 4$- $1 \over 8$ = $5 \over 8$

U₄= U₃  - $1 \over 4$U₂ =  $5 \over 8$- $3 \over 16$ = $7 \over 16$

Now, V_n= U_n+1- $1 \over 2$ U_n

V₀ = U₁- $1 \over 2$U₀=  $1 \over 2$ - $1 \over 2$(-1) =  $1 \over 2$+ $1 \over 2$= 1

V₁= U₂- $1 \over 2$U₁ =  $3 \over 4$-  $1 \over 2$( $1 \over 2$)=  $3 \over 4$ - $1 \over 4$ - $2 \over 4$

V₂ = U₃ - $1 \over 2$U₂ = $5 \over 8$- $1 \over 2$($3 \over 4$) =  $5 \over 8$ - $3 \over 8$ = $2 \over 8$

V₃ = U₄ - $1 \over 2$U₃ = $7 \over 16$ -  $1 \over 2$($5 \over 8$) =  $7 \over 16$ - $5 \over 16$= $2 \over 16$

Step 2

i)

Now, W_n =$U_n \over V_n$

W₀ = $U_0 \over V_0$ = $-1 \over 1$ = -1

W₁= $U_1 \over V_1$ = ${1 \over 2} \over { 2 \over 4}$ = 1

W₂= $U_2 \over V_2$ = ${3 \over 4} \over { 2 \over 8}$  = 3

W₃ = $U_3 \over V_3$ = ${5 \over 8} \over { 2 \over 16}$  = 5

W₀, W₁, W₂, W₃, --- = -1, 1, 3, 5, ---

We can easily observe that there is a common difference of 2 in every term.

So, it forms an arithmetic sequence with d= 2 , a = -1 where d is the difference and a the initial value.

ii) W_n = 2n - 1

for limit lim  2n - 1 = ∞

n→  ∞