Question:

Step 1: Determine the general term and the first three coefficients of the binomial expansion. The general term in the expansion of $\left(\sqrt{y} + \dfrac{1}{2\sqrt[4]{y}}\right)^n$ is given by: $$ T_{r+1} = \binom{n}{r} (\sqrt{y})^{n-r} \left(\dfrac{1}{2\sqrt[4]{y}}\right)^r $$ $$ T_{r+1} = \binom{n}{r} y^{(n-r)/2} \left(\dfrac{1}{2}\right)^r y^{-r/4} $$ $$ T_{r+1} = \binom{n}{r} \dfrac{1}{2^r} y^{\frac{n-r}{2} - \frac{r}{4}} = \binom{n}{r} \dfrac{1}{2^r} y^{\frac{2n-2r-r}{4}} = \binom{n}{r} \dfrac{1}{2^r} y^{\frac{2n-3r}{4}} $$ The coefficients of the first three terms (for $r=0, 1, 2$) are: \begin{itemize} \item For $r=0$: Coefficient is $\binom{n}{0} \dfrac{1}{2^0} = 1 \cdot 1 = 1$. \item For $r=1$: Coefficient is $\binom{n}{1} \dfrac{1}{2^1} = n \cdot \dfrac{1}{2} = \dfrac{n}{2}$. \item For $r=2$: Coefficient is $\binom{n}{2} \dfrac{1}{2^2} = \dfrac{n(n-1)}{2} \cdot \dfrac{1}{4} = \dfrac{n(n-1)}{8}$. \end{itemize} Step 2: Apply the arithmetic progression condition to find the value of $n$. The first three coefficients $1, \dfrac{n}{2}, \dfrac{n(n-1)}{8}$ form an arithmetic progression. Therefore, the middle term is the average of the first and third terms: $$ 2 \left(\dfrac{n}{2}\right) = 1 + \dfrac{n(n-1)}{8} $$ $$ n = 1 + \dfrac{n^2 - n}{8} $$ Multiply by 8 to clear the denominator: $$ 8n = 8 + n^2 - n $$ Rearrange into a quadratic equation: $$ n^2 - 9n + 8 = 0 $$ Factor the quadratic equation: $$ (n-1)(n-8) = 0 $$ This yields two possible values for $n$: $n=1$ or $n=8$. Since the expansion involves at least three terms (for coefficients up to $\binom{n}{2}$), $n$ must be greater than or equal to 2. Thus, $n=8$. Step 3: Determine the terms where the power of $y$ is a natural number. The power of $y$ in the general term $T_{r+1}$ is $\dfrac{2n - 3r}{4}$. Substitute $n=8$ into this expression: $$ \text{Power of } y = \dfrac{2(8) - 3r}{4} = \dfrac{16 - 3r}{4} $$ For the power of $y$ to be a natural number, $\dfrac{16 - 3r}{4}$ must be an integer. This implies that $16 - 3r$ must be a multiple of 4. Since $16$ is a multiple of 4, $3r$ must also be a multiple of 4. As 3 and 4 are coprime, $r$ must be a multiple of 4. The possible values for $r$ in the binomial expansion range from $0$ to $n$. With $n=8$, $r \in \{0, 1, \ldots, 8\}$. The values of $r$ that are multiples of 4 within this range are $r=0, 4, 8$. Step 4: Identify the number of terms with natural powers of $y$. Calculate the power of $y$ for each valid value of $r$: \begin{itemize} \item For $r=0$: Power of $y = \dfrac{16 - 3(0)}{4} = \dfrac{16}{4} = 4$. \item For $r=4$: Power of $y = \dfrac{16 - 3(4)}{4} = \dfrac{16 - 12}{4} = \dfrac{4}{4} = 1$. \item For $r=8$: Power of $y = \dfrac{16 - 3(8)}{4} = \dfrac{16 - 24}{4} = \dfrac{-8}{4} = -2$. \end{itemize} The powers of $y$ are $4, 1,$ and $-2$. These are all integers. Therefore, there are 3 terms in the expansion where the power of $y$ is a natural number (interpreted as an integer in this context).