two sequences , (u_n ) and (v_n ), for n∈N is defined as: { ((u_0 =3)),((u_(n+1) = (1/2)(u_n + v_n ) )) :}and { ((v_0 = 4)),((v_(n+1) = (1/2)(u_(n+1) + v_n ))) :} a) calculate u_1 ,v_1 ,u_2 and v_2 b) Another sequence (w_n ), is defined by w_n = v_n − u_n , ∀ n∈N show that w_n is a convegent geometric sequence. c) Express w_n as a function of n and obtain its limits. d) Study the sense of variation(monotony) of (u_n ) and (v_n ) what can you deduce? e) Consider another sequence t_n defined by t_n = ((u_n + 2v_n )/3) , ∀ n ∈ N show that t_n is a constant sequence f) hence obtain the limit of the sequences (u_n ) and (v


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Question Number 76630 by Rio Michael last updated on 28/Dec/19
$two sequences , (u_n ) and (v_n ), for n∈N is defined as: { ((u_0 =3)),((u_(n+1) = (1/2)(u_n + v_n ) )) :}and { ((v_0 = 4)),((v_(n+1) = (1/2)(u_(n+1) + v_n ))) :} a) calculate u_1 ,v_1 ,u_2 and v_2 b) Another sequence (w_n ), is defined by w_n = v_n − u_n , ∀ n∈N show that w_n is a convegent geometric sequence. c) Express w_n as a function of n and obtain its limits. d) Study the sense of variation(monotony) of (u_n ) and (v_n ) what can you deduce? e) Consider another sequence t_n defined by t_n = ((u_n + 2v_n )/3) , ∀ n ∈ N show that t_n is a constant sequence f) hence obtain the limit of the sequences (u_n ) and (v_n )$
$two sequences, (u_{n}) a n d (v_{n}), for n \in N is defined as :$ ${\begin{cases} u_{0} = 3 \\ u_{n + 1} = \frac{1}{2} (u_{n} + v_{n}) \end{cases} and {\begin{cases} v_{0} = 4 \\ v_{n + 1} = \frac{1}{2} (u_{n + 1} + v_{n}) \end{cases}$ $a) calculate u_{1}, v_{1}, u_{2} and v_{2}$ $b) Another sequence (w_{n}), is defined by$ $w_{n} = v_{n} - u_{n}, \forall n \in N$ $show that w_{n} is a convegent geometric sequence .$ $c) Express w_{n} as a function of n and obtain its limits .$ $d) Study the sense of variation (monotony) of (u_{n}) and (v_{n})$ $what can you deduce ?$ $e) Consider another sequence t_{n} defined by$ $t_{n} = \frac{u_{n} + 2 v_{n}}{3}, \forall n \in N$ $show that t_{n} is a constant sequence$ $f) hence obtain the limit of the sequences (u_{n}) and (v_{n})$
	Answered by mr W last updated on 29/Dec/19

	$u_{n + 1} = \frac{1}{2} (u_{n} + v_{n}) . . (i)$ $v_{n + 1} = \frac{1}{2} (u_{n + 1} + v_{n}) . . . (i i)$ $(i i) - (i) :$ $v_{n + 1} = \frac{3}{2} u_{n + 1} - \frac{1}{2} u_{n}$ $\Rightarrow v_{n} = \frac{3}{2} u_{n} - \frac{1}{2} u_{n - 1}$ $p u t t h i s i n t o (i) :$ $\Rightarrow 4 u_{n + 1} - 5 u_{n} + u_{n - 1} = 0$ $4 x^{2} - 5 x + 1 = 0$ $(4 x - 1) (x - 1) = 0$ $\Rightarrow x = \frac{1}{4}, x = 1$ $\Rightarrow u_{n} = \frac{A}{4^{n}} + B$ $u_{0} = 3 \Rightarrow 3 = A + B$ $u_{1} = \frac{1}{2} (3 + 4) = \frac{7}{2} \Rightarrow \frac{7}{2} = \frac{A}{4} + B$ $\Rightarrow - \frac{1}{2} = \frac{3 A}{4} \Rightarrow A = - \frac{2}{3}$ $\Rightarrow B = 3 + \frac{2}{3} = \frac{11}{3}$ $\Rightarrow u_{n} = \frac{1}{3} (11 - \frac{2}{4^{n}})$ $\Rightarrow v_{n} = \frac{1}{3} (11 + \frac{1}{4^{n}})$ $. . .$ $(b)$ $w_{n} = v_{n} - u_{n} = \frac{1}{3} (11 + \frac{1}{4^{n}}) - \frac{1}{3} (11 - \frac{2}{4^{n}})$ $w_{n} = \frac{1}{4^{n}} \Rightarrow G . P . w i t h c o m m o n r a t i o \frac{1}{4}$ $. . . .$ $(e)$ $t_{n} = \frac{u_{n} + 2 v_{n}}{3} = \frac{\frac{2}{3} (11 + \frac{1}{4^{n}}) + \frac{1}{3} (11 - \frac{2}{4^{n}})}{3}$ $\Rightarrow t_{n} = \frac{11}{3} = c o n s t a n t$
	Commented by mr W last updated on 29/Dec/19

	$a l t e r n a t i v e :$ $\Rightarrow 4 u_{n + 1} - 5 u_{n} + u_{n - 1} = 0$ $\Rightarrow 4 u_{n + 1} - 4 u_{n} - (u_{n} - u_{n - 1}) = 0$ $\Rightarrow (u_{n + 1} - u_{n}) = \frac{1}{4} (u_{n} - u_{n - 1})$ $\Rightarrow c_{n + 1} = \frac{1}{4} c_{n} \Rightarrow G . P .$ $\Rightarrow c_{n + 1} = c_{1} {(\frac{1}{4})}^{n}$ $c_{1} = u_{1} - u_{0} = \frac{7}{2} - 3 = \frac{1}{2}$ $\Rightarrow c_{n + 1} = \frac{1}{2} {(\frac{1}{4})}^{n}$ $\Rightarrow u_{n + 1} - u_{n} = \frac{1}{2} {(\frac{1}{4})}^{n}$ $\Rightarrow \underset{0}{\sum^{n}} u_{n + 1} - \underset{0}{\sum^{n}} u_{n} = \underset{0}{\sum^{n}} \frac{1}{2} {(\frac{1}{4})}^{n}$ $\Rightarrow u_{n + 1} - u_{0} = \frac{1}{2} \times \frac{1 - {(\frac{1}{4})}^{n + 1}}{1 - \frac{1}{4}} = \frac{2}{3} [1 - {(\frac{1}{4})}^{n + 1}]$ $\Rightarrow u_{n + 1} = \frac{2}{3} [1 - {(\frac{1}{4})}^{n + 1}] + 3 = \frac{1}{3} (11 - \frac{2}{4^{n + 1}})$ $o r$ $\Rightarrow u_{n} = \frac{1}{3} (11 - \frac{2}{4^{n}})$
	Commented by Rio Michael last updated on 29/Dec/19

	$thanks sir$
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