Summing squares and cubes without guessing
Tonight's idea
You met telescoping in grade 9. If each term of a sum is a difference, \(g(k) = f(k+1) - f(k)\), everything in the middle cancels. Writing \(\sum_{k=1}^{n}\) for "add up, for \(k\) from 1 to \(n\)":
\[\sum_{k=1}^{n} g(k) = f(n+1) - f(1).\]
So to add up \(g\), hunt for an \(f\) whose differences are \(g\). Ordinary powers are awkward: the difference of \(k^3\) is \(3k^2 + 3k + 1\). Falling powers are clean. Write \(k^{\underline{2}} = k(k-1)\) and \(k^{\underline{3}} = k(k-1)(k-2)\), each factor one less than the last. Then
\[(k+1)^{\underline{3}} - k^{\underline{3}} = k(k-1)\big[(k+1) - (k-2)\big] = 3k^{\underline{2}}.\]
The same happens for every \(m\): the difference of \(k^{\underline{m}}\) is \(m\,k^{\underline{m-1}}\). Next year you will meet the rule that the derivative of \(x^m\) is \(m\,x^{m-1}\). This is its twin, with differences in place of derivatives. Run it backwards and telescope:
\(\displaystyle\sum_{k=1}^{n} k^{\underline{m}} = \frac{(n+1)^{\underline{m+1}}}{m+1}\). For example \(\displaystyle\sum_{k=1}^{n} k(k-1) = \frac{(n+1)n(n-1)}{3}\).
Watch one: \(1^2 + 2^2 + \dots + n^2\)
Write the square in falling powers: \(k^2 = k(k-1) + k = k^{\underline{2}} + k^{\underline{1}}\). Sum each piece with the boxed rule, using \(m = 2\) and \(m = 1\):
\[\sum_{k=1}^{n} k^2 = \frac{(n+1)n(n-1)}{3} + \frac{(n+1)n}{2} = \frac{n(n+1)\big[2(n-1) + 3\big]}{6} = \frac{n(n+1)(2n+1)}{6}.\]
Check with \(n = 3\): \(1 + 4 + 9 = 14\), and \(\tfrac{3 \times 4 \times 7}{6} = 14\). Nothing was guessed. The second piece is the familiar \(1 + 2 + \dots + n = \tfrac{n(n+1)}{2}\).
Your turn
- Work out \(5^{\underline{3}} - 4^{\underline{3}}\) and check that it equals \(3 \times 4^{\underline{2}}\). Then use the formula from the example to find \(1^2 + 2^2 + \dots + 20^2\).
- Use the boxed rule to find \(\displaystyle\sum_{k=1}^{10} k(k-1)\) and \(\displaystyle\sum_{k=1}^{10} k(k-1)(k-2)\).
- Find a formula for \(1 \times 2 + 2 \times 3 + 3 \times 4 + \dots + n(n+1)\). Start by finding an \(f\) with \(f(k+1) - f(k) = k(k+1)\). Check your formula with \(n = 10\) using the sums you already know.
- Check that \(k^3 = k^{\underline{3}} + 3k^{\underline{2}} + k\). Sum each piece and simplify to show that \(1^3 + 2^3 + \dots + n^3 = \left(\tfrac{n(n+1)}{2}\right)^2\). What does that say about the sum of the first \(n\) cubes and the sum of the first \(n\) whole numbers?
Stretch: a function that is its own difference
Show that the difference of \(2^k\) is \(2^k\). Then show that \(f(k) = (k-2)2^k\) has difference \(k \cdot 2^k\), and use it to find \(1 \times 2 + 2 \times 4 + 3 \times 8 + \dots + n \cdot 2^n\).
Answers
- \(5 \times 4 \times 3 - 4 \times 3 \times 2 = 60 - 24 = 36\), and \(3 \times (4 \times 3) = 36\). \(\tfrac{20 \times 21 \times 41}{6} = 2870\).
- \(\tfrac{11 \times 10 \times 9}{3} = 330\) and \(\tfrac{11 \times 10 \times 9 \times 8}{4} = 1980\).
- \(k(k+1) = (k+1)^{\underline{2}}\), which is the difference of \(f(k) = \tfrac{(k+1)k(k-1)}{3}\). So the sum is \(f(n+1) - f(1) = \tfrac{n(n+1)(n+2)}{3}\). For \(n = 10\): \(\tfrac{10 \times 11 \times 12}{3} = 440\), and \(\sum k^2 + \sum k = 385 + 55 = 440\).
- \(k(k-1)(k-2) + 3k(k-1) + k = (k^3 - 3k^2 + 2k) + (3k^2 - 3k) + k = k^3\). Summing with \(m = 3, 2, 1\) gives \(\tfrac{(n+1)n(n-1)(n-2)}{4} + (n+1)n(n-1) + \tfrac{(n+1)n}{2}\). Take out \(\tfrac{n(n+1)}{4}\) to leave the bracket \((n-1)(n-2) + 4(n-1) + 2\). The bracket is \(n^2 + n = n(n+1)\), so the sum is \(\tfrac{n^2(n+1)^2}{4} = \left(\tfrac{n(n+1)}{2}\right)^2\). The sum of the first \(n\) cubes is the square of the sum of the first \(n\) whole numbers: \(1 + 8 + 27 = 36 = (1 + 2 + 3)^2\).
Stretch. \(2^{k+1} - 2^k = 2^k\). Next, \(f(k+1) - f(k) = (k-1)2^{k+1} - (k-2)2^k = 2^k\big(2k - 2 - k + 2\big) = k \cdot 2^k\). So the sum is \(f(n+1) - f(1) = (n-1)2^{n+1} + 2\). Check \(n = 3\): \(2 + 8 + 24 = 34 = 2 \times 16 + 2\).