Question:

Step 1: Define the general form of a term and its coefficient. The general term in the binomial expansion of $(1+2x)^n$ is given by $T_{k+1} = \binom{n}{k} (1)^{n-k} (2x)^k$. This simplifies to $T_{k+1} = \binom{n}{k} 2^k x^k$. Therefore, the coefficient of the $(k+1)$-th term is $C_{k+1} = \binom{n}{k} 2^k$. Step 2: Express the coefficients of three consecutive terms. Let the three consecutive terms be $T_r$, $T_{r+1}$, and $T_{r+2}$. Their corresponding coefficients are: Coefficient of $T_r = C_r = \binom{n}{r-1} 2^{r-1}$ Coefficient of $T_{r+1} = C_{r+1} = \binom{n}{r} 2^r$ Coefficient of $T_{r+2} = C_{r+2} = \binom{n}{r+1} 2^{r+1}$ The problem states that these coefficients are in the ratio $2:5:8$. Step 3: Set up and simplify the ratio of the first two coefficients. The ratio of $C_r$ to $C_{r+1}$ is given as $2:5$. $$ \frac{C_r}{C_{r+1}} = \frac{\binom{n}{r-1} 2^{r-1}}{\binom{n}{r} 2^r} = \frac{2}{5} $$ Using the identity $\frac{\binom{n}{k-1}}{\binom{n}{k}} = \frac{k}{n-k+1}$: $$ \frac{r}{n-(r-1)} \cdot \frac{1}{2} = \frac{r}{n-r+1} \cdot \frac{1}{2} = \frac{2}{5} $$ $$ \frac{r}{2(n-r+1)} = \frac{2}{5} $$ $$ 5r = 4(n-r+1) $$ $$ 5r = 4n - 4r + 4 $$ $$ 9r - 4n = 4 \quad \text{(Equation 1)} $$ Step 4: Set up and simplify the ratio of the second and third coefficients. The ratio of $C_{r+1}$ to $C_{r+2}$ is given as $5:8$. $$ \frac{C_{r+1}}{C_{r+2}} = \frac{\binom{n}{r} 2^r}{\binom{n}{r+1} 2^{r+1}} = \frac{5}{8} $$ Using the identity $\frac{\binom{n}{k}}{\binom{n}{k+1}} = \frac{k+1}{n-k}$: $$ \frac{r+1}{n-r} \cdot \frac{1}{2} = \frac{5}{8} $$ $$ \frac{r+1}{2(n-r)} = \frac{5}{8} $$ $$ 8(r+1) = 10(n-r) $$ $$ 8r + 8 = 10n - 10r $$ $$ 18r - 10n = -8 $$ Dividing by 2: $$ 9r - 5n = -4 \quad \text{(Equation 2)} $$ Step 5: Solve the system of linear equations for $n$ and $r$. We have two equations: 1) $9r - 4n = 4$ 2) $9r - 5n = -4$ Subtract Equation 2 from Equation 1: $$ (9r - 4n) - (9r - 5n) = 4 - (-4) $$ $$ -4n + 5n = 8 $$ $$ n = 8 $$ Substitute $n=8$ into Equation 1: $$ 9r - 4(8) = 4 $$ $$ 9r - 32 = 4 $$ $$ 9r = 36 $$ $$ r = 4 $$ Thus, $n=8$ and the terms are $T_4$, $T_5$, $T_6$. Step 6: Calculate the coefficient of the middle term. The middle term among $T_r$, $T_{r+1}$, $T_{r+2}$ is $T_{r+1}$. With $r=4$, the middle term is $T_{4+1} = T_5$. The coefficient of $T_5$ (which corresponds to $k=4$ in $T_{k+1}$) is $C_5 = \binom{n}{4} 2^4$. Substitute $n=8$: $$ C_5 = \binom{8}{4} 2^4 $$ Calculate $\binom{8}{4}$: $$ \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 $$ Calculate $2^4 = 16$. So, the middle coefficient is: $$ C_5 = 70 \times 16 = 1120 $$ The final answer is $\boxed{1120}$.