<p>If \(b_i = 1 - a_i\), \(na = \sum_{i=1}^{n} a_i\), \(nb = \sum_{i=1}^{n} b_i\), then \(\sum_{i=1}^{n} a_i b_i + \sum_{i=1}^{n} (a_i - a)^2 =\)</p>
Step-by-Step Solution
Key Concept: Since b_i = 1 - a_i, the sum ∑a_i b_i becomes ∑a_i(1 - a_i) = ∑a_i - ∑a_i². Combined with the variance term ∑(a_i - a)², which expands to ∑a_i² - na², these two expressions interact to simplify significantly.
<p><strong>Step 1:</strong> Express ∑a_i b_i using b_i = 1 - a_i:</p><p>∑a_i b_i = ∑a_i(1 - a_i) = ∑a_i - ∑a_i²</p><p><strong>Step 2:</strong> Expand ∑(a_i - a)²:</p><p>∑(a_i - a)² = ∑a_i² - 2a∑a_i + na² = ∑a_i² - 2a·na + na² = ∑a_i² - 2n·a² + na² = ∑a_i² - na²</p><p><strong>Step 3:</strong> Add the two expressions:</p><p>∑a_i b_i + ∑(a_i - a)² = (∑a_i - ∑a_i²) + (∑a_i² - na²)</p><p>= ∑a_i - na² = na - na² = na(1 - a)</p><p><strong>Step 4:</strong> Recognize that since ∑b_i = n(1 - a), we have nb = 1 - a:</p><p>∴ Answer: na·nb or equivalently na(1 - a)</p>
Correct Answer: D