<p>In the expansion of \(\left(x + \dfrac{1}{x}\right)^{13}\), every term is a function of \(x\). (State whether true or false.)</p>
Step-by-Step Solution
Key Concept: In the binomial expansion of (x + 1/x)^13, the general term is C(13,r)·x^(13-r)·(1/x)^r = C(13,r)·x^(13-2r). For a term to be independent of x (a constant), the exponent of x must equal zero, which requires 13-2r=0, giving r=6.5—a non-integer value, so NO term is independent of x.
<p><strong>Step 1:</strong> Write the general term in the expansion of (x + 1/x)^13.</p><p>T_(r+1) = C(13,r)·x^(13-r)·(1/x)^r = C(13,r)·x^(13-r-r) = C(13,r)·x^(13-2r)</p><p><strong>Step 2:</strong> Analyze the exponent of x in each term.</p><p>For r = 0, 1, 2, ..., 13, the exponent is 13-2r, which takes values: 13, 11, 9, 7, 5, 3, 1, -1, -3, -5, -7, -9, -11, -13</p><p><strong>Step 3:</strong> Verify each term contains x.</p><p>Since 13 is odd, for all valid integer values of r ∈ {0,1,2,...,13}, the exponent (13-2r) is always a non-zero integer (either positive or negative). Therefore, every term contains x with some power.</p><p><strong>Step 4:</strong> Conclusion.</p><p>Every term is of the form C(13,r)·x^k where k ≠ 0. Thus every term is a function of x.</p><p>∴ Answer: <strong>TRUE</strong> (Option A)</p>
Correct Answer: A