Sequences & Series
Sequences
Grade 11
Question:
<p>Consider the sequence 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, …. Then \(1025^{\text{th}}\) term will be</p>
<p>\(2^9\)</p>
<p>\(2^{11}\)</p>
<p>\(2^{10}\)</p>
<p>\(2^{12}\)</p>
Step-by-Step Solution
Key Concept: The sequence has n appearing exactly 2^(n-1) times. Find which 'number' contains the 1025th position by accumulating: 1 appears 1 time, 2 appears 2 times, 4 appears 4 times, 8 appears 8 times, etc. Then identify which power of 2 the 1025th term corresponds to.
<p><strong>Step 1:</strong> Identify the pattern. The number k appears 2^(k-1) times in the sequence.</p><p>- 1 appears 2^0 = 1 time (positions 1 to 1)</p><p>- 2 appears 2^1 = 2 times (positions 2 to 3)</p><p>- 4 appears 2^2 = 4 times (positions 4 to 7)</p><p>- 8 appears 2^3 = 8 times (positions 8 to 15)</p><p>- 16 appears 2^4 = 16 times (positions 16 to 31)</p><p>- 32 appears 2^5 = 32 times (positions 32 to 63)</p><p>- 64 appears 2^6 = 64 times (positions 64 to 127)</p><p>- 128 appears 2^7 = 128 times (positions 128 to 255)</p><p>- 256 appears 2^8 = 256 times (positions 256 to 511)</p><p>- 512 appears 2^9 = 512 times (positions 512 to 1023)</p><p><strong>Step 2:</strong> Calculate cumulative positions. Total positions through 512: 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 = 1023</p><p><strong>Step 3:</strong> Since position 1023 is the last occurrence of 512, position 1024 is the first occurrence of the next term (1024 = 2^10).</p><p><strong>Step 4:</strong> Position 1025 is the second occurrence of 1024.</p><p>∴ Answer: <strong>1024</strong></p>
Correct Answer: C