JEE Advanced 2025 — Paper 2 — MATH-III solution

JEE Advanced2025Paper 2MATH-III

Let $a_{0}, a_{1}, \dots, a_{23}$ be real numbers such that

$(1 + \frac{2}{5} x)^{23} = i = 0 \sum 23 a_{i} x^{i}$

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 ________.

Your answer

Solution

Step 1: Identify the ratio.

$a_{i} = (i 23) (\frac{2}{5})^{i}$ . Form the ratio

$\frac{a _{i + 1}}{a _{i}} = \frac{23 - i}{i + 1} \cdot \frac{2}{5} .$

Step 2: Find where the ratio crosses 1.

$\frac{a _{i + 1}}{a _{i}} > 1 ⟺ 2 (23 - i) > 5 (i + 1) ⟺ 46 - 2 i > 5 i + 5 ⟺ i < \frac{41}{7} \approx 5.857$ .

Step 3: Conclude.

So $a_{i + 1} > a_{i}$ for $i = 0, 1, \dots, 5$ , and $a_{i + 1} < a_{i}$ for $i \geq 6$ . The maximum is at $i = 6$ .

Therefore $r = 6$ .

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