The series follows the rule βmultiply the previous term by 3 and subtract 1β. We have 3Γ3β1 = 8, 8Γ3β1 = 23, 23Γ3β1 = 68 and 68Γ3β1 = 203. Applying the same rule again gives 203Γ3β1 = 609β1 = 608. Thus 608 is the unique value that continues the series in exact agreement with the recursive transformation.
Option A:
Option A gives 608, exactly equal to 203Γ3β1. This preserves the pattern that every term (after the first) is obtained by tripling the previous term and then subtracting 1. Because it keeps the recurrence intact without modification, 608 is the correct next term.
Option B:
Option B gives 610, which would correspond to 203Γ3+1 and change the subtraction step into an addition. This contradicts the repeated ββ1β adjustment seen at every earlier stage, so 610 is not consistent with the series.
Option C:
Option C gives 612, which would require 203Γ3+3 or another altered operation. That breaks the simple and regular rule and introduces unnecessary complexity. Therefore 612 cannot be accepted.
Option D:
Option D gives 606, implying 203Γ3β3. This again modifies the constant adjustment from β1 to β3, which is not supported by the previous transitions. Hence 606 is not a valid continuation.
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