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DanielleElmas [232]

3 years ago

9

Which sequence matches the recursive formula?

2 answers:

Ostrovityanka [42]3 years ago

8 0

A1 = 2
a2 = (a1)^2 + 3 = 2^2 + 3 = 4 + 3 = 7
a3 = (a2)^2 + 3 = 7^2 + 3 = 49 + 3 = 52
a4 = (a3)^2 + 3 = 52^2 + 3 = 2,704 + 3 = 2,707

Option C is the correct answer.

marshall27 [118]3 years ago

8 0

Answer:

The solution is 2, 7, 52, 2707, .... To find each term in the sequence, you square the previous term and add 3. This sequence is represented by the recursive formula an = (an-1)2 + 3, where a1 = 2.

So It's C.

Step-by-step explanation:

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If X²⁰¹³ + 1/X²⁰¹³ = 2, then find the value of X²⁰²² + 1/X²⁰²² = ?

enyata [817]

Step-by-step explanation:

$\bf➤ \underline{Given-} \\$

$\sf{x^{2013} + \frac{1}{x^{2013}} = 2}\\$

$\bf➤ \underline{To\: find-} \\$

$\sf {the\: value \: of \: x^{2022} + \frac{1}{x^{2022}}= ?}\\$

$\bf ➤\underline{Solution-} \\$

<u>Let us assume that:</u>

$\rm: \longmapsto u = {x}^{2013}$

<u>Therefore, the equation becomes:</u>

$\rm: \longmapsto u + \dfrac{1}{u} = 2$

$\rm: \longmapsto \dfrac{ {u}^{2} + 1}{u} = 2$

$\rm: \longmapsto{u}^{2} + 1 = 2u$

$\rm: \longmapsto{u}^{2} - 2u + 1 =0$

$\rm: \longmapsto {(u - 1)}^{2} =0$

$\rm: \longmapsto u = 1$

<u>Now substitute the value of u. We get:</u>

$\rm: \longmapsto {x}^{2013} = 1$

$\rm: \longmapsto x = 1$

<u>Therefore:</u>

$\rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 1 + 1$

$\rm: \longmapsto {x}^{2022} + \dfrac{1}{ {x}^{2022} } = 2$

★ <u>Which is our required answer.</u>

$\textsf{\large{\underline{More To Know}:}}$

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)³ = a³ + 3ab(a + b) + b³

(a - b)³ = a³ - 3ab(a - b) - b³

a³ + b³ = (a + b)(a² - ab + b²)

a³ - b³ = (a - b)(a² + ab + b²)

(x + a)(x + b) = x² + (a + b)x + ab

(x + a)(x - b) = x² + (a - b)x - ab

(x - a)(x + b) = x² - (a - b)x - ab

(x - a)(x - b) = x² - (a + b)x + ab

6 0

3 years ago