Largest Coefficient in Binomial Expansion

To find the value of $r$ for which $a_r$ is the largest among the coefficients $a_0, a_1, \dots, a_{23}$, we start by writing the general term of the binomial expansion of $\left(1 + \frac{2}{5}x\right)^{23}$.

By the Binomial Theorem: $\left(1 + \frac{2}{5}x\right)^{23} = \sum_{i=0}^{23} \binom{23}{i} \left(\frac{2}{5}\right)^i x^i$

Comparing this with the given expansion $\sum_{i=0}^{23} a_i x^i$, the coefficient $a_i$ is given by: $a_i = \binom{23}{i} \left(\frac{2}{5}\right)^i$

To determine the term with the maximum value, we examine the ratio of consecutive coefficients $\frac{a_k}{a_{k-1}}$: $\frac{a_k}{a_{k-1}} = \frac{\binom{23}{k} \left(\frac{2}{5}\right)^k}{\binom{23}{k-1} \left(\frac{2}{5}\right)^{k-1}} = \frac{\frac{23!}{k!(23-k)!}}{\frac{23!}{(k-1)!(24-k)!}} \cdot \frac{2}{5} = \frac{24-k}{k} \cdot \frac{2}{5}$

We set $\frac{a_k}{a_{k-1}} \ge 1$ to find where the sequence of coefficients is non-decreasing: $\frac{24-k}{k} \cdot \frac{2}{5} \ge 1$

$2(24 - k) \ge 5k$

$48 - 2k \ge 5k$

$7k \le 48$

$k \le \frac{48}{7} \approx 6.857$

Since $k$ must be an integer, $a_k > a_{k-1}$ for $k = 1, 2, 3, 4, 5, 6$. Specifically, for $k = 6$: $\frac{a_6}{a_5} = \frac{24 - 6}{6} \cdot \frac{2}{5} = \frac{18}{6} \cdot \frac{2}{5} = \frac{6}{5} > 1 \implies a_6 > a_5$

For $k = 7$: $\frac{a_7}{a_6} = \frac{24 - 7}{7} \cdot \frac{2}{5} = \frac{17}{7} \cdot \frac{2}{5} = \frac{34}{35} < 1 \implies a_7 < a_6$

Therefore, the terms follow the order: $a_0 < a_1 < a_2 < a_3 < a_4 < a_5 < a_6 > a_7 > a_8 > \dots > a_{23}$

Thus, $a_6$ is the largest coefficient, and the value of $r$ is $6$.