Use sigma notation and the standard results for the sums of integers, squares and cubes, and the linearity of summation, to evaluate finite series

What this dot point is asking

SEAB wants you to read and write finite sums in sigma notation, apply the standard results for the sums of the first $n$ integers, squares and cubes, use the linearity of summation, and adjust the limits of a sum when the lower limit is not $1$.

The answer

Reading sigma notation

The symbol $\displaystyle\sum_{r=1}^{n} u_r$ means "add the terms $u_r$ as $r$ runs from $1$ to $n$". The letter $r$ is a dummy variable; the lower and upper limits set the range.

The standard results

The three results you must know:

Note the neat fact that the sum of cubes is the square of the sum of integers.

Linearity

Summation distributes over addition and scalar multiples:

So any polynomial in $r$ can be summed by splitting it into the standard results.

Adjusting the limits

To sum from $r = p$ rather than $r = 1$, subtract the missing initial part:

Factorising the final expression cleanly

H2 questions almost always want a fully factorised answer, and the reliable way to get there is to pull out the common factor before expanding the bracket. When you sum a polynomial, every standard result shares the factor $\tfrac{n(n+1)}{6}$ (or a multiple of it), so factor that out first and simplify only what remains inside the bracket. In the worked example, taking out $\tfrac{n(n+1)}{6}$ left the simple bracket $(2n + 1) + 6 = 2n + 7$. Resisting the urge to multiply everything out, and instead extracting the common factor early, both saves work and produces the tidy factorised form the marks are awarded for.

Summing a geometric or telescoping series in sigma form

Not every sigma sum is a polynomial: the notation also wraps geometric series and telescoping (method-of-differences) sums. Recognising the type tells you which tool to use, the standard power results for polynomials, the geometric sum formula when the term is $ar^{k}$, and partial-fraction splitting when the term telescopes. For instance, $\sum_{r=1}^{n} \tfrac{1}{r(r+1)}$ is telescoping, not polynomial, and splits as $\tfrac{1}{r} - \tfrac{1}{r+1}$ so most terms cancel. Identifying whether the summand is polynomial, geometric, or telescoping before reaching for the standard results is the judgement that separates a confident H2 answer from a stuck one.

Examples in context

Example 1. Total of a triangular display. A shop stacks tins in rows of $1, 2, 3, \ldots, 15$. The total is $\sum_{r=1}^{15} r = \dfrac{15 \times 16}{2} = 120$ tins, a direct use of the integers result.

Example 2. Summing a quadratic cost. If the cost of producing the $r$th unit is $3r^2$ dollars, the total cost of $n$ units is $3\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{2}$ dollars, obtained by pulling the constant out and applying the squares result.

Try this

Q1. Evaluate $\displaystyle\sum_{r=1}^{n} (4r - 1)$. [2 marks]

• Cue. $4 \cdot \dfrac{n(n+1)}{2} - n = 2n(n+1) - n = 2n^2 + n$.

Q2. Express $\displaystyle\sum_{r=5}^{n} r^2$ in terms of $n$. [3 marks]

• Cue. $\dfrac{n(n+1)(2n+1)}{6} - (1 + 4 + 9 + 16) = \dfrac{n(n+1)(2n+1)}{6} - 30$.

Q3. Evaluate $\displaystyle\sum_{r=1}^{10} r^3$. [2 marks]

• Cue. $\left(\dfrac{10 \times 11}{2}\right)^2 = 55^2 = 3025$.