Sigma Notation

Writing out $3 + 7 + 11 + \dots + 79$ is clumsy, and "$\dots$" leaves the reader guessing at the pattern. Sigma notation packs the whole sum into one short expression that says exactly which terms to add. You’ll see it all through IB Mathematics: in sequences and series, in statistics formulas, in proofs by induction, and later in calculus.

Key ideas

Reading sigma notation

The capital Greek letter $\Sigma$ (sigma) means “add up”. A sum in sigma notation looks like this:

$$\sum_{r=1}^{5} (2r + 1)$$

• $r$ is the index. It starts at the number underneath ($r = 1$, the lower limit) and goes up by $1$ each time until it reaches the number on top ($r = 5$, the upper limit).
• $2r + 1$ is the general term. Substitute each value of $r$ and add the results:

$$\sum_{r=1}^{5} (2r + 1) = 3 + 5 + 7 + 9 + 11 = 35$$

The index letter doesn’t matter: $\displaystyle\sum_{r=1}^{5} (2r + 1)$ and $\displaystyle\sum_{k=1}^{5} (2k + 1)$ mean exactly the same sum. Letters like $r$, $k$, $i$ and $n$ are all common.

Counting the terms

A sum from $r = a$ to $r = b$ has

$$\text{number of terms} = b - a + 1$$

For example, $\displaystyle\sum_{r=4}^{10}$ has $10 - 4 + 1 = 7$ terms, not $6$. Forgetting the "$+ 1$" is the most common sigma mistake.

Two useful rules

Because a sigma sum is just ordinary addition, you can split it up and take out constant factors:

$$\sum_{r=1}^{n} (a\,u_r + b\,v_r) = a\sum_{r=1}^{n} u_r + b\sum_{r=1}^{n} v_r$$

And adding a constant $c$ to itself $n$ times gives $cn$:

$$\sum_{r=1}^{n} c = cn \qquad \text{for example} \qquad \sum_{r=1}^{10} 5 = 50$$

Arithmetic sums in sigma form

If the general term is linear in $r$, like $4r + 3$, the terms go up by the same amount each time, so the sum is an arithmetic series. The common difference is the coefficient of $r$. In IB notation, with first term $u_1$:

$$S_n = \frac{n}{2}\big(2u_1 + (n - 1)d\big) = \frac{n}{2}(u_1 + u_n)$$

So to evaluate the sum, find the first term, the common difference (or the last term), and the number of terms, then use the formula.

Geometric sums in sigma form

If the index is in an exponent, like $3 \times 2^r$, each term is the previous one multiplied by the same number, so the sum is a geometric series:

$$S_n = \frac{u_1(r^n - 1)}{r - 1} = \frac{u_1(1 - r^n)}{1 - r}, \qquad r \ne 1$$

Careful: here $r$ in the formula is the common ratio, which isn’t the same thing as an index called $r$. If the index letter is $r$ too, it’s safer to work out the first term and the ratio by writing out the first two or three terms.

Also watch the first term. In $\displaystyle\sum_{r=1}^{n} 3 \times 2^r$ the first term is $3 \times 2^1 = 6$, not $3$.

Changing the index

The same sum can be written with different limits. Shifting the index down by $1$ means shifting the formula up by $1$ to match:

$$\sum_{r=1}^{n} 2^{r-1} = \sum_{r=0}^{n-1} 2^{r} = 1 + 2 + 4 + \dots + 2^{n-1}$$

To change the index, substitute: if $j = r - 2$, then $r = j + 2$, so replace every $r$ with $j + 2$ and change both limits.

Using technology

Your GDC can evaluate a sigma sum directly, either with a $\Sigma$ template (type the general term and the limits) or with a list command such as sum(seq(…)). This is a great way to check an answer. In an exam, though, if you use technology you’re still expected to identify the first term and the common difference (or ratio), so write those down as part of your working.

Worked examples

Example 1: Expanding a sum

Write out the terms of $\displaystyle\sum_{r=1}^{5} (3r - 1)$ and find the sum.

Solution. Substitute $r = 1, 2, 3, 4, 5$:

$$\sum_{r=1}^{5} (3r - 1) = 2 + 5 + 8 + 11 + 14 = 40$$

Notice the terms go up by $3$, the coefficient of $r$: it’s an arithmetic series.

Example 2: Writing a sum in sigma notation

Write $4 + 9 + 14 + \dots + 99$ in sigma notation, then evaluate it.

Solution. The terms go up by $5$, so it’s arithmetic with $u_1 = 4$ and $d = 5$. The general term is

$$u_r = 4 + (r - 1)(5) = 5r - 1$$

Find which term is $99$:

$$5r - 1 = 99 \quad\Rightarrow\quad r = 20$$

So the sum is

$$4 + 9 + 14 + \dots + 99 = \sum_{r=1}^{20} (5r - 1)$$

It has $20$ terms, from $4$ to $99$:

$$S_{20} = \frac{20}{2}(4 + 99) = 10(103) = 1030$$

Check: with $r = 1$, $5(1) - 1 = 4$ ✓, and with $r = 20$, $5(20) - 1 = 99$ ✓.

Example 3: An arithmetic sum with a negative difference

Evaluate $\displaystyle\sum_{r=1}^{40} (7 - 2r)$.

Solution. Write out the first few terms to see the pattern: $5, 3, 1, -1, \dots$ It’s arithmetic with $u_1 = 5$, $d = -2$ (the coefficient of $r$), and $n = 40 - 1 + 1 = 40$ terms.

$$\begin{aligned} S_{40} &= \frac{40}{2}\big(2(5) + (40 - 1)(-2)\big) \\ &= 20(10 - 78) \\ &= -1360 \end{aligned}$$

Check with the last term: $u_{40} = 7 - 2(40) = -73$, and $\dfrac{40}{2}(5 + (-73)) = 20(-68) = -1360$ ✓.

Example 4: A geometric sum starting at zero

Evaluate $\displaystyle\sum_{k=0}^{9} 3(2)^k$.

Solution. The first few terms are $3, 6, 12, \dots$ (put in $k = 0, 1, 2$). It’s geometric with first term $u_1 = 3$ and ratio $2$.

Count the terms carefully: $k$ runs from $0$ to $9$, so there are $9 - 0 + 1 = 10$ terms.

$$S_{10} = \frac{3(2^{10} - 1)}{2 - 1} = 3(1023) = 3069$$

GDC check: enter $\displaystyle\sum_{k=0}^{9} 3(2)^k$ in the sum template; it gives $3069$ ✓.

Common mistakes

Miscounting the terms. From $r = a$ to $r = b$ there are $b - a + 1$ terms. A sum from $r = 0$ to $r = 9$ has $10$ terms. If in doubt, write out the first and last terms.

Using the coefficient as the first term of a geometric sum. In $\displaystyle\sum_{r=1}^{8} 5(0.5)^r$ the first term is $5(0.5)^1 = 2.5$, not $5$. Always substitute the lower limit to get the first term.

Treating a constant as a single term. $\displaystyle\sum_{r=1}^{10} 5$ means $5 + 5 + \dots + 5$ with ten fives, which is $50$, not $5$.

Dropping brackets. $\displaystyle\sum_{r=1}^{5} (2r + 1)$ adds $2r + 1$ each time. Without brackets, $\displaystyle\sum_{r=1}^{5} 2r + 1$ would mean “add up the $2r$‘s, then add $1$ once”, which gives a different answer. Write the brackets.

Squaring or multiplying sums term by term. $\displaystyle\sum r^2$ is not the same as $\Big(\displaystyle\sum r\Big)^2$. For $r = 1$ to $3$: $1 + 4 + 9 = 14$, but $(1 + 2 + 3)^2 = 36$. Only addition and constant factors can be split off.

Writing the GDC answer alone. If you evaluate a sum on your calculator in an exam, still show the first term, the common difference or ratio, and the number of terms.

Practice

1. (Warm-up) Write out the terms of $\displaystyle\sum_{r=1}^{4} r^2$ and find the sum.

Solution

$$\sum_{r=1}^{4} r^2 = 1 + 4 + 9 + 16 = 30$$

2. (Warm-up) How many terms does $\displaystyle\sum_{n=3}^{7} (2n + 1)$ have? Write them out and find the sum.

Solution

There are $7 - 3 + 1 = 5$ terms:

$$\sum_{n=3}^{7} (2n + 1) = 7 + 9 + 11 + 13 + 15 = 55$$

3. (Warm-up) Write $3 + 6 + 9 + \dots + 60$ in sigma notation, then evaluate it.

Solution

The terms are the multiples of $3$, so the general term is $3r$. The last term is $60 = 3(20)$, so $r$ goes from $1$ to $20$:

$$3 + 6 + 9 + \dots + 60 = \sum_{r=1}^{20} 3r$$$$S_{20} = \frac{20}{2}(3 + 60) = 10(63) = 630$$

4. (Core) Evaluate $\displaystyle\sum_{r=1}^{50} (4r + 3)$.

Solution

It’s arithmetic with $d = 4$. The first term is $4(1) + 3 = 7$ and the last term is $4(50) + 3 = 203$. There are $50$ terms.

$$S_{50} = \frac{50}{2}(7 + 203) = 25(210) = 5250$$

5. (Core) Evaluate $\displaystyle\sum_{r=1}^{8} 5(0.5)^r$. Give your answer exactly and to 3 s.f.

Solution

The first term is $5(0.5)^1 = 2.5$, the ratio is $0.5$, and there are $8$ terms. Since the ratio is less than $1$, use the second form of the formula:

$$\begin{aligned} S_8 &= \frac{2.5(1 - 0.5^8)}{1 - 0.5} \\ &= 5\left(1 - \frac{1}{256}\right) \\ &= \frac{1275}{256} \approx 4.98 \quad \text{(3 s.f.)} \end{aligned}$$

6. (Core) The series $2 - 6 + 18 - 54 + \dots$ has $8$ terms. Write it in sigma notation and find its sum.

Solution

Each term is $-3$ times the one before, so it’s geometric with $u_1 = 2$ and ratio $-3$. The general term is $2(-3)^{r-1}$:

$$2 - 6 + 18 - 54 + \dots = \sum_{r=1}^{8} 2(-3)^{r-1}$$$$S_8 = \frac{2\big((-3)^8 - 1\big)}{-3 - 1} = \frac{2(6561 - 1)}{-4} = -3280$$

7. (Core) Maya saves money each week. In week $1$ she saves $50, and each week after that she saves $5 more than the week before. Write her total savings over $26$ weeks in sigma notation, and find the total.

Solution

In week $k$ she saves $50 + 5(k - 1) = 45 + 5k$ dollars. The total over $26$ weeks, in dollars, is

$$\sum_{k=1}^{26} (45 + 5k)$$

The first term is $50$ and the last is $45 + 5(26) = 175$, so

$$S_{26} = \frac{26}{2}(50 + 175) = 13(225) = 2925$$

She saves $2925 in total.

8. (Challenge) Find the value of $n$ for which $\displaystyle\sum_{r=1}^{n} (3r + 2) = 1455$.

Solution

It’s arithmetic with first term $5$ and last term $3n + 2$, so

$$\sum_{r=1}^{n} (3r + 2) = \frac{n}{2}\big(5 + (3n + 2)\big) = \frac{n(3n + 7)}{2}$$

Set this equal to $1455$:

$$\begin{aligned} n(3n + 7) &= 2910 \\ 3n^2 + 7n - 2910 &= 0 \\ (n - 30)(3n + 97) &= 0 \end{aligned}$$

$n$ must be a positive integer, so $n = 30$.

Check: $\dfrac{30(97)}{2} = 1455$ ✓.

9. (Challenge) Rewrite $\displaystyle\sum_{r=3}^{12} (2r - 1)$ as a sum that starts at $j = 1$, using the substitution $j = r - 2$. Then evaluate it.

Solution

If $j = r - 2$, then $r = j + 2$. When $r = 3$, $j = 1$; when $r = 12$, $j = 10$. Replace $r$ in the general term:

$$2r - 1 = 2(j + 2) - 1 = 2j + 3$$$$\sum_{r=3}^{12} (2r - 1) = \sum_{j=1}^{10} (2j + 3)$$

Both give the terms $5, 7, 9, \dots, 23$. There are $10$ terms:

$$S_{10} = \frac{10}{2}(5 + 23) = 5(28) = 140$$