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-n-form-a-geome

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} + \dots . + A_{n})

A_{n} = 2 (A_{1} + A_{2} + \dots + A_{n + 1}) = 2 (A_{1} + A_{2} + \dots + 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

wow \dots