Multiple choice

Consider the sequence $1,2,2,4,4,4,4,8,8,8,8,8,8,8,8,....$ and so on. Then $1025th$ terms will be

  1. ${2}^{9}$
  2. ${2}^{11}$
  3. ${2}^{10}$
  4. ${2}^{12}$
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C Correct answer
Explanation

The sequence groups are: 2^0 (1 term), 2^1 (2 terms), 2^2 (4 terms), 2^3 (8 terms), etc. The number of terms up to group k is 1 + 2 + 4 + ... + 2^k = 2^(k+1) - 1. We want the 1025th term. 2^10 - 1 = 1023. So the 1024th term is the end of the group with value 2^9. The 1025th term starts the group with value 2^10.

AI explanation

We need to find the smallest integer k such that the sum of the powers from 0 to k is at least 1025, giving the inequality 2^{k+1} - 1 >= 1025. Solving this inequality gives 2^{k+1} >= 1026, which means k must be 10 since 2^10 is 1024 and 2^11 is 2048. The terms up to 2^9 occupy 1023 positions, meaning positions 1024 and 1025 are the first two occurrences of 2^{10}. Therefore, the 1025th term is 2^{10}.

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