Answer to Question #156269 in Algebra for wilder

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) Deduce an expression for Un in terms of n.
(ii) For every natural number n, we set Sn = Uo + U1 + U2 + ...+ Un. Show by induction that for every n E N, Sn = 2 - (2n + 3)/ 2^n.

Answer 1

Solution

• To estimate the common term of the sequence, try finding the first few terms with the help of the first relationship given

$\qquad\qquad \begin{aligned} n=0 \to u_2&=u_1-\frac{1}{4}u_0\\ u_2&=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\\ n=1\to u_3&=u_2-\frac{1}{4}u_1\\ u_3&=\frac{3}{4}-\frac{1}{8}=\frac{5}{8}\\ \end{aligned}$

• Then the sequence can be written as,

$-1,\frac{1}{2},\frac{3}{4},\frac{5}{8},.......$

• It can be rearranged to

$\frac{-1}{1},\frac{1}{2},\frac{3}{4},\frac{5}{8},.....$

• Now treating the numerator & denominator separately, common term for each can be written as follows

$\qquad\qquad \begin{aligned} U_r&=\frac{2r-1}{2^r} \end{aligned}$

• Then the common term can be written in terms of n as follows by replacing r with n

$\qquad\qquad \begin{aligned} U_n&= \frac{2n-1}{2^n} \end{aligned}$

2)

• To check with induction, first try the validity for the first possible natural numbers 0 & 1.

$\qquad\qquad \begin{aligned} S_n&=2-\frac{(2n+3)}{2^n}\\ n=0\\ S_0&= 2-3=-1=U_0\cdots(\checkmark)\\ n=1\\ S_1&= 2-\frac{5}{2}=\frac{-1}{2}=U_2\cdots(\checkmark) \end{aligned}$

• Then assume the equation holds true for an arbitrary integer $p$.

$\qquad\qquad \begin{aligned} S_p&= 2-\frac{(2p+3)}{2^p}\\ \end{aligned}$

• Now check whether the equation holds further true for integer $p+1$ ,

$\qquad\qquad \begin{aligned} S_{p+1}&=S_p+U_{p+1}\\ &=2-\frac{(2p+3)}{2^p}+\frac{2(p+1)-1}{2^{p+1}}\\ &=2-\Bigg[\frac{2(2p+3)-2(p+1)+1}{2^{p+1}}\Bigg]\\ &= 2-\Bigg[\frac{4p+6-2p-2+1}{2^{p+1}}\Bigg]\\ &= 2-\frac{(2p+5)}{2^{p+1}}\\ &\text{rearranging \, in\, terms \,of\,p+1}\\ S_{p+1}&=2-\frac{[2(p+1)+3]}{2^{p+1}}\cdots(\checkmark) \end{aligned}$

• As the equation can be further written for the $p+1$ term, it can be concluded that the equation holds true for all $n \in N$