Let $\sum_{r=0}^{200} a_r(1+x)^r = \sum_{r=0}^{200} b_r x^r$, where $a_r = 1$ ∀ $r ≥ 98$, then the greatest coefficient in the expansion of $(1+x)^{201}$ is :
Step-by-Step Solution
Key Concept: The roots of cyclotomic polynomials satisfy multiplicative relations that can be converted to additive constraints by taking real parts of power sums.
If $\lambda = a^6$ is a root of $1 + x + x^2 + \cdots + x^{10} = 0$, then $1 + \lambda + \lambda^2 + \cdots + \lambda^{10} = 0$. This gives $\text{Re}(\lambda + \lambda^2 + \cdots + \lambda^5) = -\frac{1}{2}$. Substituting $x = \mu$ in the expanded form and using that $(\mu - \beta)(\mu - \beta^2)\cdots(\mu - \beta^{10}) = 1 + \mu + \cdots + \mu^{10}$, and evaluating at $x = i$ yields the required relation.
Correct Answer: I need to analyze this problem about the greatest coefficient in an expansion.
Given: $\sum_{r=0}^{200} a_r(1+x)^r = \sum_{r=0}^{200} b_r x^r$, where $a_r = 1$ for all $r \geq 98$.
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