Summation notation (άρθρο) | Ακαδημία Khan

Summation notation (or sigma notation) allows us to write a long sum in a single expression.

Unpacking the meaning of summation notation

This is the sigma symbol:

\sum

. It tells us that we are summing something.

Let's start with a basic example:

\begin{aligned} Stop at n = 3 \\ (inclusive) \\ ↘ \\ \sum_{n = 1}^{3} & 2 n - 1 \\ ↖ \\ ↗ & Expression for each \\ Start at n = 1 & term in the sum \end{aligned}

This is a summation of the expression

2 n - 1

for integer values of

n

from

1

3

\begin{aligned} \sum_{n = 1}^{3} 2 n - 1 \\ = \underset{n = 1}{\underset{⏟}{[2 (1) - 1]}} + \underset{n = 2}{\underset{⏟}{[2 (2) - 1]}} + \underset{n = 3}{\underset{⏟}{[2 (3) - 1]}} \\ = 1 + 3 + 5 \\ = 9 \end{aligned}

Notice how we substituted

n = 1

n = 2

, and

n = 3

into

2 n - 1

and summed the resulting terms.

n

is our summation index. When we evaluate a summation expression, we keep substituting different values for our index.

Πρόβλημα 1

\sum_{n = 1}^{4} n^{2} = ?

We can start and end the summation at any value of

n

. For example, this sum takes integer values of

n

from

4

6

\begin{aligned} \sum_{n = 4}^{6} n - 1 \\ = \underset{n = 4}{\underset{⏟}{(4 - 1)}} + \underset{n = 5}{\underset{⏟}{(5 - 1)}} + \underset{n = 6}{\underset{⏟}{(6 - 1)}} \\ = 3 + 4 + 5 \\ = 12 \end{aligned}

We can use any letter we want for our index. For example, this expression has

i

for its index:

\begin{aligned} \sum_{i = 0}^{2} 3 i - 5 \\ = \underset{i = 0}{\underset{⏟}{[3 (0) - 5]}} + \underset{i = 1}{\underset{⏟}{[3 (1) - 5]}} + \underset{i = 2}{\underset{⏟}{[3 (2) - 5]}} \\ = - 5 + (- 2) + 1 \\ = - 6 \end{aligned}

Πρόβλημα 2

\sum_{k = 3}^{5} k (k + 1) =

Πρόβλημα 3

Consider the sum

4 + 25 + 64 + 121

Which expression is equal to the above sum?

Some summation expressions have variables other than the index. Consider this sum:

\sum_{n = 1}^{4} \frac{k}{n + 1}

Notice that our index is

n

, not

k

. This means we substitute the values into

n

, and

k

remains unknown:

\begin{aligned} \sum_{n = 1}^{3} \frac{k}{n + 1} \\ = \frac{k}{(1) + 1} + \frac{k}{(2) + 1} + \frac{k}{(3) + 1} \\ = \frac{k}{2} + \frac{k}{3} + \frac{k}{4} \end{aligned}

Key takeaway: Before evaluating a sum in summation notation, always make sure you identified the index, and that you are only substituting into that index. Other unknowns should remain as they are.

Πρόβλημα 4

\sum_{m = 1}^{4} 8 k - 6 m = ?

Want more practice? Try this exercise.

Μάθημα: AP®︎ Λογισμός ΑΒ > Ενότητα 6

Summation notation

Unpacking the meaning of summation notation

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