AS & A Level · Practice questions · Sums of r, r squared and r cubed

Three formulas, endless series.

Six original Cambridge-style questions on the standard results for \(\sum r\), \(\sum r^{2}\) and \(\sum r^{3}\): splitting sums, fully factorised answers, sums that do not start at 1, odd terms and finding \(n\).

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Expand, substitute, factor out, check.

Expand the general term into powers of \(r\), including any constant term.
Substitute \(\tfrac{1}{2}n(n+1)\), \(\tfrac{1}{6}n(n+1)(2n+1)\), \(\tfrac{1}{4}n^{2}(n+1)^{2}\), and \(cn\) for a constant \(c\).
Factor out common factors such as \(n(n+1)\) before expanding, and factorise what is left.
Later start: \(\sum_{r=m}^{n} = S(n) - S(m-1)\); for \(n + 1\) to \(2n\) use \(S(2n) - S(n)\).
Check your formula with \(n = 1\) or \(n = 2\); when finding \(n\), keep only a positive integer.

Recall

[2 marks]

Evaluate (a) \(\displaystyle\sum_{r=1}^{30} r\) (b) \(\displaystyle\sum_{r=1}^{12} r^{3}\).

(a) \(\tfrac{1}{2}(30)(31) = 465\). ✓
(b) \(\tfrac{1}{4}(12^{2})(13^{2}) = \left[\tfrac{1}{2}(12)(13)\right]^{2} = 78^{2} = 6084\). ✓

Calculation

[3 marks]

Show that \(\displaystyle\sum_{r=1}^{n} (6r^{2} + 2r) = 2n(n+1)^{2}\).

\(6\sum r^{2} + 2\sum r = n(n+1)(2n+1) + n(n+1)\). ✓
Take out \(n(n+1)\): \(n(n+1)\left[(2n+1) + 1\right] = n(n+1)(2n+2)\). ✓
\(2n + 2 = 2(n+1)\), so the sum is \(2n(n+1)^{2}\), as required (check \(n = 1\): \(6 + 2 = 8 = 2(1)(4)\)). ✓

Calculation

[4 marks]

Find \(\displaystyle\sum_{r=1}^{n} r(r+1)(r+2)\), giving your answer in a fully factorised form.

\(r(r+1)(r+2) = r^{3} + 3r^{2} + 2r\). ✓
Sum: \(\tfrac{1}{4}n^{2}(n+1)^{2} + \tfrac{1}{2}n(n+1)(2n+1) + n(n+1)\). ✓
Take out \(\tfrac{1}{4}n(n+1)\): \(\tfrac{1}{4}n(n+1)\left[n(n+1) + 2(2n+1) + 4\right] = \tfrac{1}{4}n(n+1)(n^{2} + 5n + 6)\). ✓
\(n^{2} + 5n + 6 = (n+2)(n+3)\), so the sum is \(\tfrac{1}{4}n(n+1)(n+2)(n+3)\) (check \(n = 1\): \(6 = \tfrac{1}{4}(24)\)). ✓

Analysis

[4 marks]

Show that \(\displaystyle\sum_{r=n+1}^{2n} r^{3} = \tfrac{1}{4}n^{2}(3n+1)(5n+3)\).

\(S(2n) = \tfrac{1}{4}(2n)^{2}(2n+1)^{2} = n^{2}(2n+1)^{2}\). ✓
The sum is \(S(2n) - S(n) = n^{2}(2n+1)^{2} - \tfrac{1}{4}n^{2}(n+1)^{2} = \tfrac{1}{4}n^{2}\left[4(2n+1)^{2} - (n+1)^{2}\right]\). ✓
The bracket is a difference of two squares: \(\left[2(2n+1) - (n+1)\right]\left[2(2n+1) + (n+1)\right] = (3n+1)(5n+3)\). ✓
So the sum is \(\tfrac{1}{4}n^{2}(3n+1)(5n+3)\), as required (check \(n = 1\): \(2^{3} = 8 = \tfrac{1}{4}(4)(8)\)). ✓

Analysis

[4 marks]

By considering \(\displaystyle\sum_{r=1}^{2n} r^{3}\) and the sum of the cubes of the even numbers, show that \(1^{3} + 3^{3} + 5^{3} + \dots + (2n-1)^{3} = n^{2}(2n^{2} - 1)\).

All cubes up to \((2n)^{3}\): \(\tfrac{1}{4}(2n)^{2}(2n+1)^{2} = n^{2}(2n+1)^{2}\). ✓
Even cubes: \(\displaystyle\sum_{r=1}^{n} (2r)^{3} = 8 \cdot \tfrac{1}{4}n^{2}(n+1)^{2} = 2n^{2}(n+1)^{2}\). ✓
Odd cubes \(=\) all minus even \(= n^{2}\left[(2n+1)^{2} - 2(n+1)^{2}\right] = n^{2}\left[4n^{2} + 4n + 1 - 2n^{2} - 4n - 2\right]\). ✓
\(= n^{2}(2n^{2} - 1)\), as required (check \(n = 2\): \(1 + 27 = 28 = 4 \times 7\)). ✓

Analysis

[5 marks]

(a) Show that \(\displaystyle\sum_{r=1}^{n} r(r+2) = \tfrac{1}{6}n(n+1)(2n+7)\).

(b) Hence find the least value of \(n\) for which \(\displaystyle\sum_{r=1}^{n} r(r+2) > 10\,000\).

(a) \(\sum r^{2} + 2\sum r = \tfrac{1}{6}n(n+1)(2n+1) + n(n+1) = \tfrac{1}{6}n(n+1)\left[(2n+1) + 6\right] = \tfrac{1}{6}n(n+1)(2n+7)\). ✓
(b) The sum increases with \(n\). Roughly \(\tfrac{1}{3}n^{3} \approx 10\,000\), so \(n \approx 31\); try values near this. ✓
\(n = 29\): \(\tfrac{1}{6}(29)(30)(65) = 9425 < 10\,000\). ✓
\(n = 30\): \(\tfrac{1}{6}(30)(31)(67) = 10\,385 > 10\,000\). ✓
The least value is \(n = 30\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.