3. Summation Formulae

Video coming soon2 worked examples

Theory

Standard results (given on the formula sheet) let us sum powers of the counting variable:

\sum_{r=1}^{n} r = \frac{n(n+1)}{2}, \quad \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{r=1}^{n} r^3 = \left[\frac{n(n+1)}{2}\right]^2

Summation is linear, so a sum of several terms can be split apart, and a constant $c$ sums to:

\sum_{r=1}^{n} c = cn

The Golden Rule: split the summation using linearity, apply each standard formula, then factorise — the answer almost always simplifies to a neat factorised form.

⚠️ Common Examiner Traps

Summing a constant: $\sum_{r=1}^{n} c = cn$ , not $c$ .
Coefficients: a factor multiplies the whole standard formula, e.g. $\sum 2r = 2 \cdot \frac{n(n+1)}{2}$ .
Not simplifying: the exam usually wants the fully factorised expression, not an unexpanded sum of fractions.

Worked examples

Example 1

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

Step 1: Split the summation using linearity:

\sum_{r=1}^{n} (2r+1) = 2\sum_{r=1}^{n} r + \sum_{r=1}^{n} 1

Step 2: Apply the standard results:

= 2 \cdot \frac{n(n+1)}{2} + n = n(n+1) + n

Step 3: Simplify and factorise:

= n^2 + n + n = n^2 + 2n = n(n+2)

Example 2

Find a formula for $\displaystyle\sum_{r=1}^{n} r(r + 2)$ in fully factorised form.

Step 1: Expand and split:

\sum_{r=1}^{n} r(r+2) = \sum_{r=1}^{n} (r^2 + 2r) = \sum_{r=1}^{n} r^2 + 2\sum_{r=1}^{n} r

Step 2: Apply the standard results:

= \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{6} + n(n+1)

Step 3: Take out the common factor $\dfrac{n(n+1)}{6}$ and simplify:

= \frac{n(n+1)}{6}\bigl[(2n+1) + 6\bigr] = \frac{n(n+1)(2n+7)}{6}