Sequences & Series
Sequence and Series
Allen Star Batch
Grade 11
Question:
Let $a_i + b_i = 1\forall i = 1,2,...,6$ and $a = \frac{1}{6}(a_1 + a_2 + ... + a_6), b = \frac{1}{6}(b_1 + b_2 + ... + b_6)$. Then $a_0b_1 + a_1b_2 + ... + a_6b_0 = nab - (a_1 - a_2)^2 - (a_2 - a_3)^2 - ... - (a_6 - a_1)^2$ where $n$ is equal to
Step-by-Step Solution
Key Concept: Expanding squared terms and using the mean constraint collapses the expression to a simple product form.
Given $a + b = \frac{1}{6}\sum_{i=1}^{6}(a_i + b_i) = 1$, we compute $\sum_{i=1}^{6}(a_i - a)^2 + \sum_{i=1}^{6}a_i b_i = 6a^2 - 2a\sum a_i + \sum a_i^2 + \sum a_i b_i$. Using $\sum a_i = 6a$ and the constraint, this simplifies to $6a^2 - 2a(6a) + \sum a_i^2 + \sum a_i b_i = -6a^2 + 6a + 6a(1-a) = 6ab$.
Correct Answer: 6