Binomial Theorem
Integer and fractional parts
Grade 11
Question:
<p>If \((4 + \sqrt{15})^n = I + f\), where \(n\) is an odd natural number, \(I\) is an integer and \(0 < f < 1\), then</p>
<p>(1) \(I\) is an odd integer</p>
<p>(2) \(I\) is an even integer</p>
<p>(3) \((I+f)(I-f) = 1\)</p>
<p>(4) \(I - f = (4 - \sqrt{15})^n\)</p>
Step-by-Step Solution
Key Concept: Since (4 + √15)^n + (4 - √15)^n always yields an even integer (by binomial expansion, irrational terms cancel), and (4 - √15)^n is a small positive number less than 1 for odd n, we can write I + f + (4 - √15)^n = even integer, giving f + (4 - √15)^n = 1.
<p><strong>Step 1:</strong> Note that 4 - √15 ≈ 0.127, so 0 < (4 - √15)^n < 1 for any positive n.</p><p><strong>Step 2:</strong> By binomial expansion, (4 + √15)^n + (4 - √15)^n produces only rational terms (irrational parts cancel), yielding an even integer. Let this sum = 2k where k is an integer.</p><p><strong>Step 3:</strong> Therefore: (4 + √15)^n + (4 - √15)^n = 2k, which gives: I + f + (4 - √15)^n = 2k</p><p><strong>Step 4:</strong> Since I is an integer, f must be of the form f = m - (4 - √15)^n where m is an integer. Given 0 < f < 1, we have m = 1.</p><p><strong>Step 5:</strong> Thus f + (4 - √15)^n = 1, so:</p><p>(A) f(1 - f) = f · (4 - √15)^n ✓ (evaluates to positive value)</p><p>(C) [f] = 0 (integer part of f is 0) ✓</p><p>(D) 1 - f = (4 - √15)^n ✓</p><p>∴ Answer: ACD</p>
Correct Answer: ACD