<p>If \((1^2 - t_1) + (2^2 - t_2) + \cdots + (n^2 - t_n) = \dfrac{n(n^2 - 1)}{3}\), then \(t_n\) is equal to</p>
Step-by-Step Solution
Key Concept: Sum the left side algebraically by separating squared terms from t_i terms, then equate coefficients with the right side to isolate the pattern for t_n.
<p><strong>Step 1:</strong> Expand the left side by separating terms:</p><p>∑(i=1 to n)[i² - tᵢ] = ∑(i=1 to n)i² - ∑(i=1 to n)tᵢ</p><p><strong>Step 2:</strong> Use the formula ∑(i=1 to n)i² = n(n+1)(2n+1)/6:</p><p>n(n+1)(2n+1)/6 - ∑(i=1 to n)tᵢ = n(n²-1)/3</p><p><strong>Step 3:</strong> Simplify the right side: n(n²-1)/3 = n(n-1)(n+1)/3</p><p><strong>Step 4:</strong> Solve for the sum of t terms:</p><p>∑(i=1 to n)tᵢ = n(n+1)(2n+1)/6 - n(n-1)(n+1)/3</p><p>= n(n+1)[2n+1 - 2(n-1)]/6 = n(n+1)(3)/6 = n(n+1)/2</p><p><strong>Step 5:</strong> Recognize that ∑(i=1 to n)tᵢ = n(n+1)/2 is the sum of first n natural numbers, so tₙ = n</p><p>∴ Answer: B</p>
Correct Answer: B