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Sagot :
Let's address the problem of finding the first two terms of the sequence defined recursively by the formula [tex]\( a_n = (a_{n-1})^2 + 4 \)[/tex], given that [tex]\( a_4 = 1604 \)[/tex].
To find the earlier terms, we'll work backwards from [tex]\( a_4 \)[/tex] to [tex]\( a_3 \)[/tex] and then to [tex]\( a_2 \)[/tex].
Step 1: Finding [tex]\( a_3 \)[/tex]
The recursive formula gives us:
[tex]\[ a_4 = (a_3)^2 + 4 \][/tex]
Substitute the known value:
[tex]\[ 1604 = (a_3)^2 + 4 \][/tex]
Solving for [tex]\( a_3 \)[/tex]:
[tex]\[ (a_3)^2 = 1600 \][/tex]
[tex]\[ a_3 = \sqrt{1600} \][/tex]
[tex]\[ a_3 = 40.0 \][/tex]
Step 2: Finding [tex]\( a_2 \)[/tex]
Again using the recursive formula:
[tex]\[ a_3 = (a_2)^2 + 4 \][/tex]
Substitute the known value of [tex]\( a_3 \)[/tex]:
[tex]\[ 40 = (a_2)^2 + 4 \][/tex]
Solving for [tex]\( a_2 \)[/tex]:
[tex]\[ (a_2)^2 = 36 \][/tex]
[tex]\[ a_2 = \sqrt{36} \][/tex]
[tex]\[ a_2 = 6.0 \][/tex]
Therefore, the first two terms of the sequence are:
[tex]\[ a_2 = 6.0 \][/tex]
[tex]\[ a_3 = 40.0 \][/tex]
Thus, the correct answer is:
6, 40
To find the earlier terms, we'll work backwards from [tex]\( a_4 \)[/tex] to [tex]\( a_3 \)[/tex] and then to [tex]\( a_2 \)[/tex].
Step 1: Finding [tex]\( a_3 \)[/tex]
The recursive formula gives us:
[tex]\[ a_4 = (a_3)^2 + 4 \][/tex]
Substitute the known value:
[tex]\[ 1604 = (a_3)^2 + 4 \][/tex]
Solving for [tex]\( a_3 \)[/tex]:
[tex]\[ (a_3)^2 = 1600 \][/tex]
[tex]\[ a_3 = \sqrt{1600} \][/tex]
[tex]\[ a_3 = 40.0 \][/tex]
Step 2: Finding [tex]\( a_2 \)[/tex]
Again using the recursive formula:
[tex]\[ a_3 = (a_2)^2 + 4 \][/tex]
Substitute the known value of [tex]\( a_3 \)[/tex]:
[tex]\[ 40 = (a_2)^2 + 4 \][/tex]
Solving for [tex]\( a_2 \)[/tex]:
[tex]\[ (a_2)^2 = 36 \][/tex]
[tex]\[ a_2 = \sqrt{36} \][/tex]
[tex]\[ a_2 = 6.0 \][/tex]
Therefore, the first two terms of the sequence are:
[tex]\[ a_2 = 6.0 \][/tex]
[tex]\[ a_3 = 40.0 \][/tex]
Thus, the correct answer is:
6, 40
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