Let $a_0, a_1, \ldots, a_{23}$ be real numbers such that $\left(1 + \frac{2}{5}x\right)^{23} = \sum_{i=0}^{23} a_i x^i$ for every real number $x$. Let $a_r$ be the largest among the numbers $a_j$ for $0 \le j \le 23$. Then the value of $r$ is ________.

Let a₀, a₁, …, a₂₃ be real numbers such that (1 + 2/5x)²³ = ∑ᵢ₌₀²³ aᵢ xⁱ for every real number x. Let aᵣ be the largest among the numbers aⱼ for 0 ≤ j ≤ 23. Then the value of r is ________. $\left(1 + \frac{2}{5}x\right)^{23} = \sum_{i=0}^{23} a_i x^i$ $a_0, a_1, \ldots, a_{23} \in \mathbb{R}$ $\text{To find: } r \text{ such that } a_r = \max\{a_j : 0 \le j \le 23\}$ $1.\ \text{general coefficient } a_i \text{ from binomial theorem}$ $2.\ \text{ratio } a_{i+1}/a_i \text{ and where it crosses } 1$ $3.\ \text{the } i \text{ at the peak is } r$ $(1+y)^n = \sum_{i=0}^{n} \binom{n}{i} y^i$ $a_i = \binom{23}{i}\left(\frac{2}{5}\right)^i$ $\frac{a_{i+1}}{a_i} = \frac{\binom{23}{i+1}}{\binom{23}{i}}\cdot\frac{2}{5} = \frac{23-i}{i+1}\cdot\frac{2}{5}$ $\frac{23-i}{i+1}\cdot\frac{2}{5} > 1 \implies 2(23-i) > 5(i+1)$ $46 - 2i > 5i + 5 \implies 41 > 7i \implies i < \frac{41}{7} \approx 5.86$ $i \le 5:\ a_{i+1} > a_i \ \Rightarrow\ a_6 > a_5$ $i \ge 6:\ a_{i+1} < a_i \ \Rightarrow\ a_6 > a_7$ $i=5:\ \frac{18}{6}\cdot\frac{2}{5} = 1.2 > 1$ $i=6:\ \frac{17}{7}\cdot\frac{2}{5} \approx 0.97 < 1$ $a_r = \max_j a_j \ \Rightarrow\ r = 6$ $\boxed{r = 6}$