<p>If \(S_n\) denotes the sum of \(n\) terms of A.P., then \(S_{n+3} - 3S_{n+2} + 3S_{n+1} - S_n =\)</p>
Step-by-Step Solution
Key Concept: Apply the finite difference operator repeatedly on S_n: the first difference of an A.P. sum gives the n-th term, and successive differences of terms in A.P. (which form another A.P.) eventually vanish. Here, three successive differences of S_n yields 0 because we're differencing a quadratic function three times.
<p><strong>Step 1:</strong> Recall that for an A.P., S_n = An² + Bn (where A and B are constants depending on first term and common difference).</p><p><strong>Step 2:</strong> Define the finite difference operator: ΔS_n = S_{n+1} - S_n = a_{n+1} (the (n+1)-th term).</p><p><strong>Step 3:</strong> The second difference: Δ²S_n = Δ(S_{n+1} - S_n) = a_{n+2} - a_{n+1} = d (constant, the common difference).</p><p><strong>Step 4:</strong> The third difference: Δ³S_n = Δ(a_{n+2} - a_{n+1}) = d - d = 0.</p><p><strong>Step 5:</strong> The given expression S_{n+3} - 3S_{n+2} + 3S_{n+1} - S_n represents the third finite difference Δ³S_n (using binomial coefficients: (1,-3,3,-1) are the binomial expansion coefficients of (1-1)³).</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: D