Which one of the following is the difference of the sum of cubes of first ten natural numbers and the sum of squares of first ten natural numbers?

Correct — B, 2640. Two standard formulas settle this in one line. The sum of the first n cubes is [n(n+1)/2]², so for n = 10 it is (10 × 11 / 2)² = 55² = 3025. The sum of the first n squares is n(n+1)(2n+1)/6, so for n = 10 it is (10 × 11 × 21)/6 = 385. The difference is 3025 − 385 = 2640. Adding the ten terms by hand reaches the same place: 1³ − 1², 2³ − 2², and so on is 0 + 4 + 18 + 48 + 100 + 180 + 294 + 448 + 648 + 900, which totals 2640.

• (a)2400 — Subtract 2400 from the sum of cubes and the sum of squares would have to be 625. The sum of the squares of 1 to 10 is 385; 625 is simply 25 squared.
• (c)2880 — This needs the sum of squares to be 145, which no partial sum of 1², 2², 3² … reaches at the tenth term — the running totals go 1, 5, 14, 30, 55, 91, 140, 204, 285, 385.
• (d)2000 — A round guess. It would require the sum of squares to be 1025, nearly three times its actual value of 385.

Both sums have closed forms that are worth carrying into the hall. The sum of the first n natural numbers is n(n+1)/2; the sum of their squares is n(n+1)(2n+1)/6; and the sum of their cubes is the square of the first of these, [n(n+1)/2]². That last identity is the elegant part — the cubes of 1 to n always add up to the square of the sum of 1 to n.

The question is engineered so that manual addition wastes the two minutes it is worth. Ten cubes are 1, 8, 27, 64, 125, 216, 343, 512, 729 and 1000, and adding those under exam pressure invites a slip; the formula gives 3025 straight away. The second discipline is checking an offered answer backwards. Since the sum of cubes is fixed at 3025, every option implies a particular value for the sum of squares, and only one of those values is 385. That reverse check takes seconds and rules out three options at once.

• The sum of the first n cubes equals [n(n+1)/2]², the square of the sum of the first n natural numbers.
• The sum of the first n squares equals n(n+1)(2n+1)/6.
• For n = 10 the two sums are 3025 and 385, and 3025 − 385 = 2640.
• Term by term the same difference is the sum of n²(n − 1) from 1 to 10, that is 0 + 4 + 18 + 48 + 100 + 180 + 294 + 448 + 648 + 900.
• Both formulas are worth memorising for n = 10 alone, since 55, 385 and 3025 recur across series questions.

• Adding ten cubes by hand when a one-line formula is available.
• Applying the square formula to the cubes, or the other way round.
• Computing 10 × 11 ÷ 2 = 55 for the cubes and then forgetting to square it.

Asked as a pure formula-recall item — the examiner is testing whether the two summation formulas are at your fingertips, not your addition.

• practice — not a real PYQ

  What is the sum of the cubes of the first five natural numbers?

  • (a)225
  • (b)125
  • (c)100
  • (d)275

  Answer(a) 225 — [5 × 6 ÷ 2]² = 15² = 225, which is also 1 + 8 + 27 + 64 + 125.

• practice — not a real PYQ

  The sum of the squares of the first n natural numbers is 285. What is n?

  • (a)8
  • (b)9
  • (c)10
  • (d)11

  Answer(b) 9 — n(n+1)(2n+1)/6 with n = 9 gives (9 × 10 × 19) ÷ 6 = 285.