Question

Let $ a_0, a_1, ..., 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 \leq j \leq 23 $. Then the value of $ r $ is ________.

Hint:

To find the maximum term in a binomial-type expansion with a coefficient \( r^i \), use the approximation: \[ \text{Maximum at } i = \left\lfloor \frac{(n + 1)r}{1 + r} \right\rfloor \] for \( a_i = \binom{n}{i} r^i \)

Answer 1

Step 1: Understanding the Problem

We are given that:

\[(1 + \frac{2}{5}x)^{23} = \sum_{i=0}^{23} a_i x^i\]

This means that \(a_i\) represents the coefficient of \(x^i\) in the expansion of \((1 + \frac{2}{5}x)^{23}\).

Step 2: General Term of the Expansion

From the binomial theorem, the general term is:

\[T_{i} = \binom{23}{i} \left(\frac{2}{5}x\right)^i\]

So, the coefficient of \(x^i\) is:

\[a_i = \binom{23}{i}\left(\frac{2}{5}\right)^i\]

Step 3: Behavior of Coefficients

We need to find which \(a_i\) is the largest among \(a_0, a_1, \dots, a_{23}\).

To do this, consider the ratio:

\[\frac{a_{i+1}}{a_i} = \frac{\binom{23}{i+1}\left(\frac{2}{5}\right)^{i+1}}{\binom{23}{i}\left(\frac{2}{5}\right)^i}\]

Simplify:

\[\frac{a_{i+1}}{a_i} = \frac{23-i}{i+1} \cdot \frac{2}{5}\]

Step 4: Condition for Increasing Sequence

If \(\frac{a_{i+1}}{a_i} > 1\), the sequence increases; if less than 1, it decreases.

\[\frac{23-i}{i+1} \cdot \frac{2}{5} > 1\]

\[\frac{23-i}{i+1} > \frac{5}{2}\]

\[2(23-i) > 5(i+1)\]

\[46 - 2i > 5i + 5\]

\[46 - 5 > 7i\]

\[41 > 7i\]

\[i < \frac{41}{7} \approx 5.857\]

Step 5: Identifying the Maximum

Thus, the coefficients increase up to \(i = 5\) and then start decreasing after \(i = 6\)

Therefore, the largest coefficient occurs at \(i = 6\).

Final Answer

The value of \(r\) is:

r = 6