Binomial Theorem
Binomial Theorem
star_batch_jee_advanced_2025
Grade 11
Question:
In the expansion of $(x + a)^n, n \in \mathbb{N}$, if the sum of odd numbered terms be $\alpha$ and the sum of even numbered terms be $\beta$, then:
$4\alpha\beta = (x + a)^{2n} - (x - a)^{2n}$
$2(\alpha^2 + \beta^2) = (x + a)^{2n} + (x - a)^{2n}$
$\alpha^2 - \beta^2 = (x^2 - a^2)^n$
$\alpha^2 + \beta^2 = (x + a)^{2n} + (x - a)^{2n}$
Step-by-Step Solution
Key Concept: The key is recognizing that $(x+a)^n$ and $(x-a)^n$ differ only in the signs of even-positioned terms, allowing us to isolate $\alpha$ and $\beta$ through addition and subtraction.
In the expansion of $(x+a)^n$, we have $(x+a)^n = \sum_{k=0}^{n} \binom{n}{k}x^{n-k}a^k$. The odd-positioned terms (1st, 3rd, 5th...) sum to $\alpha$ and even-positioned terms (2nd, 4th, 6th...) sum to $\beta$. Notice that $(x+a)^n = \alpha + \beta$ and $(x-a)^n = \alpha - \beta$ (since odd-positioned terms remain same while even-positioned terms change sign). From these: $\alpha = \frac{(x+a)^n + (x-a)^n}{2}$ and $\beta = \frac{(x+a)^n - (x-a)^n}{2}$. Computing $\alpha^2 - \beta^2 = (\alpha+\beta)(\alpha-\beta) = (x+a)^n \cdot (x-a)^n = [(x+a)(x-a)]^n = (x^2-a^2)^n$.
Correct Answer: 3