Suppose the nth term of a series is 1 + n/2 + n²/2. If there are 20 terms in the series, then the sum of the series is equal to

Correct — D, 1560. Break the sum into three pieces, because a sum of terms is the sum of the pieces. The constant 1 appears in every one of the twenty terms and contributes 20. The n/2 piece contributes half the sum of the first twenty natural numbers, which is (20 × 21 ÷ 2) ÷ 2 = 210 ÷ 2 = 105. The n²/2 piece contributes half the sum of the first twenty squares, and the sum of the first n squares is n(n+1)(2n+1)/6, giving (20 × 21 × 41) ÷ 6 = 2,870, of which half is 1,435. Adding, 20 + 105 + 1,435 = 1,560.

• (a)1360 — Two hundred short. It is what you reach if the sum of squares is taken as 2,470 instead of 2,870, a slip in the triple product 20 × 21 × 41.
• (b)1450 — Close to the sum of the square terms alone, 1,435, with a small addition. It comes from dropping either the constant or the linear part of the general term.
• (c)1500 — A round number that no partial combination of 20, 105 and 1,435 produces. It is the kind of value offered to catch an approximate calculation.

Three summation formulas cover almost every series question of this type. 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, [n(n+1)/2]². Summation is linear, so a general term that is a combination of constants, n and n² can be summed piece by piece and the constants pulled outside.

For n = 20 the three values worth writing down first are 210, 2,870 and, if cubes appear, 44,100. Once they are on the page the arithmetic is only halving and adding. A quick sanity check is available too: the largest term, at n = 20, is 1 + 10 + 200 = 211, and the smallest, at n = 1, is 2, so twenty terms averaging somewhere near seventy-eight should total in the high hundreds to mid-fifteen hundreds — enough to rule out 1,360 on inspection if the working has gone astray.

• Sum of the first n natural numbers = n(n+1)/2; for n = 20 this is 210.
• Sum of the first n squares = n(n+1)(2n+1)/6; for n = 20 this is 2,870.
• Summation is linear, so a general term can be summed piece by piece.
• The three contributions here are 20, 105 and 1,435.
• The twentieth term itself is 1 + 10 + 200 = 211.

• Summing only the n² part and forgetting the constant and linear pieces.
• Applying the cube formula where the square formula is needed.
• Slipping in the triple product 20 × 21 × 41, which is where most arithmetic errors on this item occur.

A direct application of the summation formulas. The examiner is checking whether the formulas are available instantly, not whether you can add twenty terms.

• practice — not a real PYQ

  The sum of the squares of the first 10 natural numbers is

  • (a)285
  • (b)385
  • (c)485
  • (d)555

  Answer(b) 385 — from n(n+1)(2n+1)/6 with n = 10.

• practice — not a real PYQ

  If the nth term of a series is 2n + 3, the sum of its first 10 terms is

  • (a)110
  • (b)120
  • (c)130
  • (d)140

  Answer(c) 130 — 2 × 55 for the n part plus 30 for the constant.