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Sagot :
The first five terms of the recursively defined sequence are 6, 12, 48, 768, 196608
For given question,
We have been given the recursive formula of a sequence.
[tex]a_{k+1}=\frac{1}{3}{a_k}^2[/tex]
Also, the first term of the sequence is,
a1 = 6
Substitute k = 1 in given recursive formula.
⇒ [tex]a_{1+1}=\frac{1}{3}{a_1}^2[/tex]
⇒ a2 = 1/3 (6)²
⇒ a2 = (1/3) × 36
⇒ a2 = 12
Substitute k = 2 in given recursive formula.
⇒ [tex]a_{2+1}=\frac{1}{3}{a_2}^2[/tex]
⇒ a3 = (1/3) × (12)²
⇒ a3 = (1/3) × 144
⇒ a3 = 48
Substitute k = 3 in given recursive formula.
⇒ [tex]a_{3+1}=\frac{1}{3}{a_3}^2[/tex]
⇒ a4 = (1/3) × (48)²
⇒ a4 = (1/3) × 2304
⇒ a4 = 768
Substitute k = 4 in given recursive formula.
⇒ [tex]a_{4+1}=\frac{1}{3}{a_4}^2[/tex]
⇒ a5 = (1/3) × (768)²
⇒ a5 = (1/3) × 589824
⇒ a5 = 196608
Therefore, the first five terms of the recursively defined sequence are 6, 12, 48, 768, 196608
Learn more about the recursive formula of sequence here:
https://brainly.com/question/14457800
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