Question 10

The coefficient $a_i$ in the expansion of $(1 + \frac{2}{5}x)^{23}$ is given by $a_i = \binom{23}{i} \left(\frac{2}{5}\right)^i$.

To find the largest coefficient $a_r$, we check the ratio $\frac{a_r}{a_{r-1}} \ge 1$:

$\frac{\binom{23}{r} (2/5)^r}{\binom{23}{r-1} (2/5)^{r-1}} \ge 1 \implies \frac{23-r+1}{r} \cdot \frac{2}{5} \ge 1 \implies 2(24-r) \ge 5r \implies 48 - 2r \ge 5r \implies 7r \le 48 \implies r \le 6.85$.

Now check $\frac{a_{r+1}}{a_r} \le 1$:

$\frac{\binom{23}{r+1} (2/5)^{r+1}}{\binom{23}{r} (2/5)^r} \le 1 \implies \frac{23-r}{r+1} \cdot \frac{2}{5} \le 1 \implies 46 - 2r \le 5r + 5 \implies 7r \ge 41 \implies r \ge 5.85$.

Since $r$ must be an integer, $r = 6$.