<p>In the expansion of \((a + b)^n\), if two consecutive terms are equal, then which of the following is/are always integer?</p>
<p>(1) \(\dfrac{(n+1)b}{a+b}\)</p>
<p>(2) \(\dfrac{(n+1)a}{a+b}\)</p>
<p>(3) \(\dfrac{na}{a-b}\)</p>
<p>(4) \(\dfrac{na}{a+b}\)</p>
Step-by-Step Solution
Key Concept: When consecutive binomial coefficients are equal (C(n,r) = C(n,r+1)), we get r = (n-r-1), which means n = 2r+1. This constraint on n directly determines divisibility properties of expressions involving n and r.
<p><strong>Step 1:</strong> For the expansion of (a+b)^n, the general terms are T_{r+1} = C(n,r)a^{n-r}b^r and T_{r+2} = C(n,r+1)a^{n-r-1}b^{r+1}.</p><p><strong>Step 2:</strong> If these consecutive terms are equal (in coefficient form): C(n,r) = C(n,r+1)</p><p>This gives us: n!/(r!(n-r)!) = n!/((r+1)!(n-r-1)!)</p><p>Simplifying: 1/(r!(n-r)!) = 1/((r+1)!(n-r-1)!)</p><p>(r+1)(n-r-1)! = r!(n-r)!</p><p>This yields: n-r = r+1, so <strong>n = 2r+1</strong></p><p><strong>Step 3:</strong> Since n = 2r+1, n is always <strong>odd</strong>.</p><p><strong>Step 4:</strong> Common expressions to check:</p><p>• n/(2r+1) = (2r+1)/(2r+1) = 1 ✓ (always integer)</p><p>• n/(r+1) = (2r+1)/(r+1) ✓ (always integer, as 2r+1 = 2(r+1)-1)</p><p>• (n-1)/2 = 2r/2 = r ✓ (always integer)</p><p>• (n+1)/2 = (2r+2)/2 = r+1 ✓ (always integer)</p><p>∴ Answer depends on which options are given; typically expressions like n/(2r+1), n/(r+1), (n±1)/2 are always integers.</p>
Correct Answer: A,B