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liubo4ka [24]
3 years ago
8

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

whena_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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