Sigma Notation
A single Greek letter compresses an entire sum into a formula — reading it correctly is the first step to evaluating it fast.
Reading sigma notation
The capital Greek letter $\Sigma$ ("sigma") is shorthand for "sum of." The expression
means: substitute $r=1$ into $f(r)$, then $r=2$, then $r=3$, and so on up to $r=n$, adding every result together. The number below $\Sigma$ (here, $r=1$) is the lower limit — the starting value of $r$ — and the number above (here, $n$) is the upper limit, the final value of $r$. The expression $f(r)$ itself is the general term, describing the rule for each term in the sum. Written in full,
The limits don't have to start at 1 — a sum like $\sum_{r=5}^{10} f(r)$ simply runs $r$ from 5 to 10 inclusive, giving 6 terms in total.
If the general term is just the constant $1$ (not depending on $r$ at all), then $\displaystyle\sum_{r=1}^{n} 1 = \underbrace{1+1+\cdots+1}_{n\text{ times}} = n$. More generally, $\sum_{r=1}^{n} c = cn$ for any constant $c$ — "1" is simply added (or any constant added) once for every value $r$ takes, and there are $n$ such values.
Combining sigma notation with series results
Sigma notation often disguises an arithmetic or geometric series in a compact form. Two properties make evaluating such sums easier:
Sums split over addition/subtraction: $\displaystyle\sum_{r=1}^{n}\big[f(r)\pm g(r)\big] = \sum_{r=1}^{n}f(r) \pm \sum_{r=1}^{n}g(r)$. Constants factor out: $\displaystyle\sum_{r=1}^{n} k\,f(r) = k\sum_{r=1}^{n}f(r)$ for any constant $k$. Together, these let a sum like $\sum_{r=1}^{n}(3r+2)$ be broken into $3\sum r + \sum 2$, each evaluated with a standard formula from the previous two lessons.
Recognising the general term's structure is key: if $f(r)$ is linear in $r$ (like $3r+2$), the sum is really an arithmetic series in disguise, and $\frac{n}{2}[2a+(n-1)d]$ or $\sum r=\frac{n(n+1)}{2}$ applies directly. If $f(r)$ involves $r$ as an exponent (like $2^{r}$ or $3(2)^{r-1}$), it's a geometric series, and $\frac{a(1-r^n)}{1-r}$ applies instead.
Worked example 1
Writing out the first few terms explicitly — even when a formula could shortcut the work — is a reliable way to confirm the general term and limits have been read correctly before committing to further algebra.
Worked example 2
This sum is exactly the arithmetic series with $u_r=3r+2$, so $u_1=5$, $d=3$, $n=20$ terms, giving $S_{20}=\frac{20}{2}[2(5)+19(3)]=10[10+57]=670$ — matching perfectly, and confirming both approaches are equivalent.
Worked example 3
Spotting whether $r$ appears as a multiplier (arithmetic, e.g. $3r+2$) or as an exponent (geometric, e.g. $3(2)^{r-1}$) in the general term is the single most useful skill for sigma-notation questions — it immediately signals which sum formula to reach for.
Worked example 4
When the lower limit isn't 1, the number of terms is $(\text{upper}-\text{lower}+1)$, not the upper limit itself — a common source of off-by-one errors. Method B here, treating the sum directly as an arithmetic series with the correct term count, is often faster and safer than splitting into two sums from 1.
Assuming a sum always starts at $r=1$ and miscounting the number of terms when the lower limit is shifted — for $\sum_{r=6}^{15}$, there are $15-6+1=10$ terms, not $15$ or $9$. Always compute (upper limit − lower limit + 1) explicitly before applying a sum formula.
Check yourself
Evaluate $\displaystyle\sum_{r=1}^{12}(5r-3)$.
(Hint: split the sum using the standard results for $\sum r$ and $\sum 1$.)