<p>In the binomial expansion of \(\left(\sqrt{y} + \dfrac{1}{2\sqrt[4]{y}}\right)^n\) the first three coefficients form an arithmetic progression. Then:</p>
<p>(a) the value of \(n\) is 7</p>
<p>(b) the value of \(n\) is 8</p>
<p>(c) number of terms in the expansion where the power of \(y\) is natural is 2</p>
<p>(d) number of terms in the expansion where the power of \(y\) is natural is 3</p>
Step-by-Step Solution
Key Concept: The first three binomial coefficients C(n,0), C(n,1), C(n,2) form an AP when 2·C(n,1) = C(n,0) + C(n,2). This gives a quadratic equation in n that determines the expansion uniquely.
<p><strong>Step 1:</strong> Identify the first three binomial coefficients in the expansion of $\left(\sqrt{y} + \frac{1}{2\sqrt[4]{y}}\right)^n$.</p><p>These are: $C(n,0) = 1$, $C(n,1) = n$, $C(n,2) = \frac{n(n-1)}{2}$</p><p><strong>Step 2:</strong> Apply the AP condition: $2 \cdot C(n,1) = C(n,0) + C(n,2)$</p><p>$$2n = 1 + \frac{n(n-1)}{2}$$</p><p>$$4n = 2 + n(n-1)$$</p><p>$$4n = 2 + n^2 - n$$</p><p>$$n^2 - 5n + 2 = 0$$</p><p><strong>Step 3:</strong> Wait, this doesn't give integer solutions. Reconsider: the first three <em>terms</em> have coefficients $C(n,0)$, $C(n,1)\left(\frac{1}{2}\right)$, $C(n,2)\left(\frac{1}{2}\right)^2$.</p><p>For AP of actual coefficients: $2 \cdot n \cdot \frac{1}{2} = 1 + \frac{n(n-1)}{2} \cdot \frac{1}{4}$</p><p>$$n = 1 + \frac{n(n-1)}{8}$$</p><p>$$8n = 8 + n^2 - n$$</p><p>$$n^2 - 9n + 8 = 0$$</p><p>$$(n-1)(n-8) = 0$$</p><p>Since $n > 2$, we have $n = 8$.</p><p><strong>Step 4:</strong> With $n=8$, the general term is $T_{r+1} = C(8,r)(\sqrt{y})^{8-r}\left(\frac{1}{2\sqrt[4]{y}}\right)^r = C(8,r)\frac{1}{2^r}y^{\frac{8-3r}{4}}$</p><p>For rational powers: $\frac{8-3r}{4}$ must be an integer, so $8-3r \equiv 0 \pmod{4}$, giving $r \in \{0, 4, 8\}$.</p><p>∴ Answer: <strong>BD</strong></p>
Correct Answer: BD