A_1 ,A_2 ,A_3 ,…,A_n are defined as follows A_1 =2, A_(n−1) =2(A_1 +A_2 +A_3 +…+A_n ) for n=1,2,3,… (a)Evaluate the numerical values of A_(3 ) and A_4 (b)Prove that A_2 ,A_3 ,…,A


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Question Number 122999 by ZiYangLee last updated on 21/Nov/20

$A_{1}, A_{2}, A_{3}, \dots, A_{n} are defined as follows$ $A_{1} = 2, A_{n - 1} = 2 (A_{1} + A_{2} + A_{3} + \dots + A_{n})$ $f o r n = 1, 2, 3, \dots$ $(a) Evaluate the numerical values of A_{3} and A_{4}$ $(b) Prove that A_{2}, A_{3}, \dots, A_{n} form a geometric$ $progression .$
	Answered by mathmax by abdo last updated on 21/Nov/20

	$A_{1} = 2 and A_{2} = 2 (A_{1} + A_{2} + A_{3})$ $A_{1} = 2 (A_{1} + A_{2}) \Rightarrow - A_{1} = {2 A}_{2} \Rightarrow A_{2} = - \frac{1}{2} A_{1} = - 1$ $A_{2} = {2 A}_{1} + {2 A}_{2} + {2 A}_{3} \Rightarrow {2 A}_{3} = - A_{2} - {2 A}_{1} = 1 - 4 = - 3 \Rightarrow$ $A_{3} = - \frac{3}{2}$ $A_{3} = 2 (A_{1} + A_{2} + A_{3} + A_{4}) = {2 A}_{1} + {2 A}_{2} + {2 A}_{3} + {2 A}_{4} \Rightarrow$ ${2 A}_{4} = A_{3} - {2 A}_{1} - {2 A}_{2} - {2 A}_{3} = - A_{3} - {2 A}_{1} - {2 A}_{2}$ $= \frac{3}{2} - 4 + 2 = \frac{3}{2} - 2 = - \frac{1}{2} \Rightarrow A_{4} = - \frac{1}{4}$ $\frac{A_{3}}{A_{2}} = - \frac{3}{2} \times (- 1) = \frac{3}{2} \Rightarrow \frac{A_{2}}{A_{3}} = \frac{2}{3}$ $\frac{A_{4}}{A_{3}} = - \frac{1}{4} \times \frac{- 2}{3} = \frac{2}{3}$ $let suppose A_{n} geometric complex \Rightarrow \exists q \in C / A_{n - 1} = {qA}_{n}$ $= 2 (A_{1} + A_{2} + . . . . + A_{n})$ $A_{n} = 2 (A_{1} + A_{2} + . . . + A_{n + 1}) = 2 (A_{1} + A_{2} + . . . + A_{n}) + {2 A}_{n + 1}$ $= A_{n - 1} + {2 A}_{n + 1} = q A_{n} + {2 A}_{n + 1} \Rightarrow (1 - q) A_{n} = {2 A}_{n + 1} \Rightarrow$ $A_{n} = \frac{2}{1 - q} A_{n + 1} \Rightarrow \frac{2}{1 - q} = q \Rightarrow 2 = q - q^{2} \Rightarrow - q^{2} + q - 2 = 0 \Rightarrow$ $q^{2} - q + 2 = 0 if A_{n} complex we get$ $Δ = 1 - 8 = - 7 \Rightarrow q_{1} = \frac{1 + i \sqrt{7}}{2} and q_{2} = \frac{1 - i \sqrt{7}}{2}$
	Commented by ZiYangLee last updated on 21/Nov/20

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