Binomial Theorem
PYP_JEE_ADV_2025_P2
Grade None
Question:
Let $a_0, a_1, \dots, a_{23}$ be real numbers such that
$$\left(1 + \dfrac{2}{5}x\right)^{23} = \sum_{i=0}^{23} a_i x^i$$
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
Step-by-Step Solution
Key Concept: Finding the greatest term in a binomial expansion by analyzing the ratio of successive terms $a_{i+1}/a_i$.
By binomial theorem, the coefficient $a_i$ of $x^i$ in the expansion of $\left(1 + \dfrac{2}{5}x\right)^{23}$ is:
$$a_i = \binom{23}{i} \left(\dfrac{2}{5}\right)^i$$
Consider the ratio $\dfrac{a_{i+1}}{a_i}$:
$$\dfrac{a_{i+1}}{a_i} = \dfrac{\binom{23}{i+1}\left(\dfrac{2}{5}\right)^{i+1}}{\binom{23}{i}\left(\dfrac{2}{5}\right)^i} = \dfrac{23-i}{i+1} \cdot \dfrac{2}{5}$$
Set $\dfrac{a_{i+1}}{a_i} > 1$:
$$\dfrac{23-i}{i+1} \cdot \dfrac{2}{5} > 1 \implies 46 - 2i > 5i + 5 \implies 41 > 7i \implies i < \dfrac{41}{7} \approx 5.857$$
Since $i$ is an integer, this holds for $i \le 5$. Thus:
$$a_0 < a_1 < a_2 < a_3 < a_4 < a_5 < a_6$$
For $i \ge 6$, the ratio $\dfrac{a_{i+1}}{a_i} < 1 \implies a_6 > a_7 > a_8 > \dots$
Thus, the maximum value is $a_6$, which means $r = 6$.
Correct Answer: