Special sums of natural numbers

Three standard sums appear so often that you should know them by heart: $\sum_{k=1}^n k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2.$ Combined with the linearity of $\Sigma$, they let you sum any series whose general term is a polynomial in $k$.

Derivations

Sum of first $n$ natural numbers. Use the AP formula with $a = 1, d = 1, n$ terms: $\sum_{k=1}^n k = \frac{n}{2}(1 + n) = \frac{n(n+1)}{2}.$ Pictorially: pair $1$ with $n$, $2$ with $n - 1$, $\dots$ , each pair sums to $n + 1$, and there are $n/2$ pairs.

Sum of squares. A standard derivation uses the identity $(k + 1)^3 - k^3 = 3k^2 + 3k + 1$. Telescope: $(n + 1)^3 - 1 = \sum_{k=1}^n [(k+1)^3 - k^3] = 3 \sum k^2 + 3 \sum k + n.$ Solve for $\sum k^2$: $3 \sum k^2 = (n + 1)^3 - 1 - 3 \cdot \frac{n(n+1)}{2} - n,$ $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n + 1)}{6}.$

Sum of cubes. Similarly, expand $(k+1)^4 - k^4 = 4 k^3 + 6 k^2 + 4 k + 1$ and telescope: $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2 = \left(\sum_{k=1}^n k\right)^2.$

The remarkable fact: the sum of the first $n$ cubes equals the square of the sum of the first $n$ integers.

Using these formulas

To sum any polynomial series $\sum_{k=1}^n p(k)$ where $p$ has degree $\le 3$, write $p(k)$ as a linear combination of $1, k, k^2, k^3$ and apply linearity. For higher polynomial degrees, extend with $\sum k^4 = \dfrac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30}$ (a Class XI bonus, rarely required).

Worked examples

Example 1. Compute $\sum_{k=1}^{20} k$.

$\dfrac{20 \cdot 21}{2} = 210$.

Example 2. Compute $\sum_{k=1}^{15} k^2$.

$\dfrac{15 \cdot 16 \cdot 31}{6} = \dfrac{7440}{6} = 1240$.

Example 3. Find $\sum_{k=1}^n (k + 2)^2$.

Expand: $\sum (k^2 + 4k + 4) = \sum k^2 + 4 \sum k + 4n = \dfrac{n(n+1)(2n+1)}{6} + 2 n(n+1) + 4n$.

For $n = 5$: $55 + 30 \cdot 2 + 20 = 55 + 60 + 20 = 135$. (Check: $3^2 + 4^2 + 5^2 + 6^2 + 7^2 = 9 + 16 + 25 + 36 + 49 = 135$. ✓)

Example 4. Find $\sum_{k=1}^n k(k + 1)(k + 2)$.

$k(k+1)(k+2) = k^3 + 3 k^2 + 2k$. So $\sum = \sum k^3 + 3 \sum k^2 + 2 \sum k = \left(\frac{n(n+1)}{2}\right)^2 + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}.$ This simplifies to $\dfrac{n(n+1)(n+2)(n+3)}{4}$.

Example 5. Find the sum $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n(n+1)$.

$T_k = k(k + 1) = k^2 + k$. Sum $= \sum k^2 + \sum k = \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n+1)}{2} = \dfrac{n(n+1)(n+2)}{3}$.

Try it yourself

1. Compute $1 + 2 + 3 + \dots + 100$.
2. Compute $1^2 + 2^2 + 3^2 + \dots + 50^2$.
3. Compute $1^3 + 2^3 + 3^3 + \dots + 20^3$.
4. Show that $\sum_{k=1}^n (2k - 1) = n^2$. (Sum of first $n$ odd numbers.)
5. Find $\sum_{k=1}^n (2k - 1)^2$.
6. Find $\sum_{k=1}^n k(2k - 1)$.
7. Find $\sum_{k=1}^{n} (k + 1)(k + 2)$.
8. Find $\sum_{k=1}^n k(k + 2)$.
9. Compute $\sum_{k=1}^{15} (k^2 + k)$.
10. Find $\sum_{k=1}^n (n - k + 1) k$.
11. Show $1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$.
12. Find $\sum_{k=1}^n k(n - k)$.

Pitfalls / Tricks

• Memorise the three sums; they are unavoidable.
• Always expand the general term first, then split into known sums.
• $\sum k^3 = (\sum k)^2$ is a beautiful identity , useful for cross-checks.
• Insight. A polynomial in $k$ summed to $n$ produces a polynomial in $n$ of degree one higher , this is the basic shape of every result.