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3 years ago

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What are the first four terms of the sequence defined recursively below

2 answers:

Effectus [21]3 years ago

4 0

At n = 2
a₂ = (-1)² + 3a₂₋₁ + 2
a₂ = 1 + 3a₁ + 2
a₂ = 1 + (3)(5) + 2
a₂ = 18

At n = 3
a₃ = (-1)³ + 3a₃₋₁ + 2
a₃ = -1 + 3a₂ + 2
a₃ = -1 + (3)(18) + 2
a₃ = 55

At n = 4
a₄ = (-1)⁴ + 3a₄₋₁ + 2
a₄ = 1 + 3a₃ + 2
a₄ = 1 + (3)(55) + 2
a₄ = 168

The first four terms are 5, 18, 55, 168

NikAS [45]3 years ago

4 0

Answer:

The first four terms are 5, 18, 55, 168

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Apply perfect square formula: (a-b)² = a²-2ab+b²

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$x=\sqrt{57}+6$

also solving

$x-6=-\sqrt{57}$

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$x-6+6=-\sqrt{57}+6$

Simplify

$x=-\sqrt{57}+6$

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