4.4 Sigma notation for sums — Learn

Prerequisites: generating terms of a sequence from a formula in Section 4.3; expanding and simplifying algebraic expressions; the laws of indices.

By the end of this section you can: read a sum written with sigma notation; write a sum of terms in sigma notation; evaluate a sum term by term; and use the standard results for $\sum r$, $\sum r^2$ and $\sum r^3$ to add many terms at once.

1. Sigma notation

The capital Greek letter sigma, $\Sigma$, means "add up". It saves writing a long list of terms.

Here $r$ is the index. It is a counter, not a quantity in the problem. The number below the sigma is the first value of the index, and the number above is the last. The expression after the sigma is the summand: the rule for each term.

Example 1. Evaluate $\sum_{r=1}^{5}(3r + 1)$.

Write out the terms by substituting $r = 1, 2, 3, 4, 5$:

The letter $r$ disappears when the terms are written out. It was only a counter, so any letter would do.

2. Writing a sum in sigma notation

To write a list of terms as a sigma sum, find the rule that produces them and count how many there are.

Example 2. Write $2 + 4 + 6 + \dots + 20$ in sigma notation.

The terms are the even numbers. The rule for the $r$th even number is $2r$. The terms run from $r = 1$ (giving $2$) to $r = 10$ (giving $20$). So

Checking the last term is the way to confirm the upper limit: $2 \times 10 = 20$, which is the term given.

3. Working with the index

Two rules let you split a sum or take out a factor:

for a constant $k$. So

A constant inside the sum is added once for each value of the index, which gives

4. The standard results

Four standard results turn a long sum into one calculation. Each one starts at $r = 1$ and stops at $r = n$.

Example 3. Use the standard results to work out $\sum_{r=1}^{20}(4r^2 - 2r)$.

Split the sum and take out each factor:

Use the two standard results with $n = 20$:

So

Adding twenty terms by hand would take much longer, and one slip would spoil the answer.

5. When the index does not start at 1

The standard results start at $r = 1$. To start later, add the missing terms and take them away again.

Example 4. Use the standard results to work out $\sum_{r=5}^{10} r$.

Check by adding the terms: $5 + 6 + 7 + 8 + 9 + 10 = 45$. The two methods agree, so the adjustment is correct.

Watch out. The index limits are part of the sum, and they are often misread.

Using a standard result from the wrong start. The formula $\sum_{r=1}^{n}r = \frac{n(n+1)}{2}$ assumes the sum begins at $r = 1$. For $\sum_{r=5}^{10} r$ the answer $55$ is the sum from $1$ to $10$, not from $5$ to $10$. Re-check: add the first few terms by hand. For $\sum_{r=5}^{10}r$ the first term is $5$, so the total must be less than $55$.

Mixing up the term and the sum. The expression $u_n$ is one term, the $n$th. The sigma sum is the total of many terms. For the sequence $u_n = 50n + 100$, the term $u_{12}$ is $700$, but the sum of the first twelve terms is $5100$.

Progress check

Return to the chapter problem. The quiz night raises $50n + 100$ pounds in month $n$, and the online appeal raises $600(0.8)^{n-1}$ pounds.

1. Write the total raised by the quiz night in the first 12 months in sigma notation.
2. Use the standard result for $\sum n$ to work out that total.
3. Write the total raised by the appeal in the first 12 months in sigma notation.
4. Explain why the standard result for $\sum r$ cannot be used for the appeal sum.

You should now be able to

• read the index and the limits of a sigma sum;
• write a list of terms in sigma notation, and check the last term;
• split a sum and take a constant factor outside it;
• use the standard results for $\sum r$, $\sum r^2$ and $\sum r^3$;
• adjust a sum whose index does not start at $1$;
• tell the difference between a single term and a sum of terms.