JEE Advanced 2025 Paper 2 · Q10 · Binomial Theorem

Question

Let $a_0, a_1, \ldots, a_{23}$ 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 $0 \le j \le 23$.

Then the value of $r$ is __________.

SolveKar AI · Accepted Answer

$a_j = \binom{23}{j} \left(\frac{2}{5}ight)^j$. The ratio $\frac{a_{j+1}}{a_j} = \frac{23 - j}{j + 1} \cdot \frac{2}{5}$. The sequence increases as long as this ratio $> 1$, i.e., $2(23 - j) > 5(j + 1)$ $	o$ $46 - 2j > 5j + 5$ $	o$ $j < \frac{41}{7} \approx 5.857$. So the largest term occurs at $j = 6$.

JEE Advanced 2025 Paper 2 · Q10 · Binomial Theorem

Solution