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A sequence of positive integers a1 , a2,...

A sequence of positive integers `a_1 , a_2, a_3, …..a_n` is given by the rule `a_(n + 1) = 2a_(n) + 1`. The only even number in the sequence is 38. What is the value of `a_2`?

A

`11`

B

`25`

C

`38`

D

`77`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the value of \( a_2 \) in the sequence defined by the recurrence relation \( a_{n+1} = 2a_n + 1 \), given that the only even number in the sequence is 38. ### Step-by-Step Solution: 1. **Identify the Sequence Rule**: The sequence is defined by the recurrence relation: \[ a_{n+1} = 2a_n + 1 \] This means each term is generated from the previous term by doubling it and adding 1. 2. **Determine the Position of the Even Number**: Since 38 is the only even number in the sequence, it must be the first term \( a_1 \). This is because all subsequent terms generated from an odd number (which is the case for all terms after the first) will also be odd. 3. **Set \( a_1 \)**: Therefore, we set: \[ a_1 = 38 \] 4. **Calculate \( a_2 \)**: Using the recurrence relation, we can find \( a_2 \): \[ a_2 = 2a_1 + 1 \] Substituting \( a_1 = 38 \): \[ a_2 = 2 \times 38 + 1 \] \[ a_2 = 76 + 1 = 77 \] 5. **Conclusion**: Thus, the value of \( a_2 \) is: \[ a_2 = 77 \] ### Final Answer: The value of \( a_2 \) is **77**. ---
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