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3 years ago

What are the first two terms of the sequence defined recursively by the formula

Mathematics

2 answers:

taurus [48]3 years ago

6 0

Answer: C. $\sqrt2,\ 6$

Step-by-step explanation:

Given formula $a_n=(a_{n-1})^2+4$

when $a_4=1604$

Substitute n=4 in th given formula , we get

$a_4=(a_{4-1})^2+4\\\Rightarrow1604=(a_3)^2+4\\\Rightarrow(a_3)^2=1604-4\\\Rightarrow(a_3)^2=1600\\\Rightarrow\ a_3=40$

Substitute n=3 in the given formula, we get

$a_3=(a_{3-1})^2+4\\\Rightarrow40=(a_2)^2+4\\\Rightarrow(a_2)^2=40-4\\\Rightarrow(a_2)^2=36\\\Rightarrow\ a_2=6$

Substitute n=2 in the given formula, we get

$a_2=(a_{2-1})^2+4\\\Rightarrow6=(a_1)^2+4\\\Rightarrow(a_1)^2=6-4\\\Rightarrow(a_1)^2=2\\\Rightarrow\ a_1=\sqrt{2}$

Thus, the first two terms of the sequence= $\sqrt{2}\ and \ 6$

MrRa [10]3 years ago

5 0

A3^2+4=1604

a3=40

a2^2+4=40

a2=6

a1^2+4=6

a1=√2

So the first two terms are √2 and 6, answer C

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Previous concepts

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Solution to the problem

Part a

For this case we can use the binomial model since we assume independent events and the same probability for each trial is the same p =0.15

Part b

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And replacing we got:

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Part d

For this cae we want thi probability:

$P(X \geq 1)$

And we can use the complment rule and we got:

$P(X \geq 1)= 1-P(X$

And using the result from part a we got:

$P(X \geq 1)= 1-P(X$

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