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.

Also, the first term of the sequence is,
a1 = 6
Substitute k = 1 in given recursive formula.
⇒ 
⇒ a2 = 1/3 (6)²
⇒ a2 = (1/3) × 36
⇒ a2 = 12
Substitute k = 2 in given recursive formula.
⇒ 
⇒ a3 = (1/3) × (12)²
⇒ a3 = (1/3) × 144
⇒ a3 = 48
Substitute k = 3 in given recursive formula.
⇒ 
⇒ a4 = (1/3) × (48)²
⇒ a4 = (1/3) × 2304
⇒ a4 = 768
Substitute k = 4 in given recursive formula.
⇒ 
⇒ 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:
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Answer:
120÷5=24 / 24×4=96 / 96+24=120
Step-by-step explanation:
Answer:
Hope this helps sorry I'm in class right now
Step-by-step explanation:
The left hand side expression of the given equation is a difference of two squares. The first term, x², is a square of x and the second term, 25 is the square of 5. The factors of the expression are (x - 5) and (x + 5).
(x - 5)(x + 5) = 0
The values of x from the equation above are x = -5 and x = 5.
The answer is C..............