Step-by-step Derivations

1. Identify the General Coefficient $a_i$

The given binomial expansion is: $\left(1+\frac{2}{5} x\right)^{23}=\sum\limits_{i=0}^{23} a_i x^i$ Using the binomial theorem, the expansion of $(1+y)^n$ is $\sum_{i=0}^{n} \binom{n}{i} y^i$. In our case, $n=23$ and $y = \frac{2}{5}x$. So, the expansion is: $\left(1+\frac{2}{5} x\right)^{23} = \sum_{i=0}^{23} \binom{23}{i} \left(\frac{2}{5}x\right)^i = \sum_{i=0}^{23} \binom{23}{i} \left(\frac{2}{5}\right)^i x^i$ By comparing the coefficients of $x^i$ with the given expression, we find the general form of the coefficient $a_i$: $a_i = \binom{23}{i} \left(\frac{2}{5}\right)^i$

2. Condition for the Largest Coefficient

To find the index $r$ for which the coefficient $a_r$ is the largest, we need to find the value of $r$ where the sequence of coefficients $a_0, a_1, a_2, \ldots$ stops increasing and starts decreasing. This means that $a_r$ must be greater than or equal to its adjacent coefficients, $a_{r-1}$ and $a_{r+1}$. The key condition to analyze is the ratio of consecutive coefficients: $\frac{a_r}{a_{r-1}} \ge 1$ This inequality will tell us for which values of $r$ the coefficients are increasing or have reached their maximum.

3. Calculate the Ratio of Consecutive Coefficients

Let's compute the ratio $\frac{a_r}{a_{r-1}}$ for $r \ge 1$: $\frac{a_r}{a_{r-1}} = \frac{\binom{23}{r} \left(\frac{2}{5}\right)^r}{\binom{23}{r-1} \left(\frac{2}{5}\right)^{r-1}}$ We use the property $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$. Here, $n=23$ and $k=r$. $\frac{a_r}{a_{r-1}} = \frac{23-r+1}{r} \cdot \frac{2}{5} = \frac{24-r}{r} \cdot \frac{2}{5}$

4. Solve the Inequality

Now, we apply the condition from Step 2: $\frac{24-r}{r} \cdot \frac{2}{5} \ge 1$ Since $r$ is an index in the sum, $r \ge 1$, so $r$ is positive. We can multiply both sides by $5r$ without changing the inequality direction: $2(24-r) \ge 5r$ $48 - 2r \ge 5r$ $48 \ge 7r$ $r \le \frac{48}{7}$ $r \le 6.857...$

5. Interpret the Result

The inequality $a_r/a_{r-1} \ge 1$ holds for integers $r=1, 2, 3, 4, 5, 6$. This implies: $a_1 > a_0, a_2 > a_1, \ldots, a_6 > a_5$ So, the sequence of coefficients is increasing up to $a_6$: $a_0 < a_1 < a_2 < a_3 < a_4 < a_5 < a_6$ For $r > 6.857...$, which means for $r=7, 8, \ldots, 23$, the inequality is reversed, i.e., $a_r/a_{r-1} < 1$. This means $a_r < a_{r-1}$. For $r=7$, we have $a_7 < a_6$. For $r=8$, we have $a_8 < a_7$, and so on. $a_6 > a_7 > a_8 > \ldots > a_{23}$ Combining both results, we see that $a_6$ is the largest coefficient in the sequence. Therefore, the value of $r$ for the largest coefficient is 6.

Final Answer

The value of $r$ is 6.