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2 years ago

write the first five terms of the sequence defined by the recursive formula an = 2 (an-1)^2, with a1 = 1.

Mathematics

1 answer:

vredina [299]2 years ago

3 0

Answer:

1, 2, 8, 128, 32768

Step-by-step explanation:

Given

The recursive formula aₙ = 2*(aₙ₋₁)², with a₁ = 1

The first 5 terms are:

a₁ = 1
a₂ = 2*1² = 2
a₃ = 2*2² = 8
a₄= 2*8² = 128
a₅ = 2*128² = 32768

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$\large\boxed{\text{sixth term is equal to}\ \dfrac{1}{1024}}$

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<h2>It's not a geometric sequence.</h2>

If $a_3=-\dfrac{1}{16}$ then the common ratio is $r=\dfrac{-\frac{1}{16}}{\frac{1}{4}}=-\dfrac{1}{16}\cdot\dfrac{4}{1}=-\dfrac{1}{4}$

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$a_n=-1\left(-\dfrac{1}{4}\right)^{n-1}$

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$-1\left(-\dfrac{1}{4}\right)^{n-1}=\dfrac{1}{1024}\qquad\text{use}\ a^n:a^m=a^{n-m}\\\\-\left(-\dfrac{1}{4}\right)^n:\left(-\dfrac{1}{4}\right)^1=\dfrac{1}{1024}\\\\-\left(-\dfrac{1}{4}\right)^n\cdot(-4)=\dfrac{1}{1024}\\\\(4)\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{1024}\qquad\text{divide both sides by 4}\ \text{/multiply both sides by}\ \dfrac{1}{4}/\\\\\left(-\dfrac{1}{4}\right)^n=\dfrac{1}{4096}\\\\\dfrac{(-1)^n}{4^n}=\dfrac{1}{4^6}\qquad n\ \text{must be even number. Therefore}\ (-1)^n=1$

$\dfrac{1}{4^n}=\dfrac{1}{4^6}\iff n=6$

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