Question

Let Sn = 1 + 3 + 11 + 25 + 45 + … . Then sum up to 20th term equals, do not provide me a solution but give me the most efficient method by which I should approach this question to solve it quickly during an exam

Answer 1

Use the method of expressing terms in a polynomial in n, find the general term Tn, then sum via known formulas for n, n², n³.

Answer 2

Efficient Exam Strategy

1. Observe the sequence differences
   • Compute ΔTn = Tn – Tn–1: 3–1=2, 11–3=8, 25–11=14, 45–25=20,…
   • These form an arithmetic sequence with first term 2 and common difference 6.
2. Conclude Tn is quadratic
   • Since second differences are constant (6), Tn must be of form an² + bn + c.
3. Find coefficients a, b, c quickly
   • Use T1=1, T2=3, and the second difference =2a=6 ⇒ a=3.
   • Solve for b, c in T1: 3·1² + b·1 + c =1 and T2: 3·4 + 2b + c =3.
4. Write general term
   • Obtain Tn=3n² – 3n + 1.
5. Use sum formulas
   • Sn=∑(3n²) – ∑(3n) + ∑1 = 3·[n(n+1)(2n+1)/6] – 3·[n(n+1)/2] + n.
6. Plug n=20
   • Compute each part with minimal arithmetic in exam.

This approach avoids term-by-term addition and leverages standard power‑sum formulas for a quick result.