Summation symbol (Sigma Notation)

\[ \text{$ \sum_{\color{red}{i=1}} ^{\color{red}{3}}{\color{#1299e7}{\textit{i}}} = \color{#1299e7}{1+2+3}=6$}\]

How do we do that? We plug in into i (that is next to a summation sign) values starting from the lower bound, (here 1) up to the upper bound incrementing by one. And we sum all these numbers. That's all!

Let's look on the summation symbol in all its glory and purpose!
If you had to express the sum of the first 1000 natural numbers, what would you choose?

Read in Ukrainian

Let's solve more similar examples:

Example 1:

$ \sum_{\color{red}{i=1}} ^{\color{red}{5}}{\color{#1299e7}{\textit{i}}} = ?$

Solution: we simply plug in numbers for i from 1 to 5 and sum them: $ \sum_{i=1} ^{5}{{\textit{i}}} =1+2+3+4+5=15$

Solution: we simply plug in numbers for i from 1 to 5 and sum them: $ \sum_{i=1} ^{5}{{\textit{i}}} =1+2+3+4+5=\\=15$

Example 2:

$ \sum_{\color{red}{i=1}} ^{\color{red}{3}}{\color{#1299e7}{2\cdot i}} =?$

Solution: multiply each number from 1 to 3 by 2 and sum them: $ \sum_{i=1} ^{3}{\color{blue}{2}\cdot i} =\color{blue}{2}\cdot1+\color{blue}{2}\cdot2+\color{blue}{2}\cdot3=12$

Solution: multiply each number from 1 to 3 by 2 and sum them: $ \sum_{i=1} ^{3}{\color{blue}{2}\cdot i} =\color{blue}{2}\cdot1+\color{blue}{2}\cdot2+\\+\color{blue}{2}\cdot3=12$

What if the summation index does not start from 1? In such cases, we simply plug in the values from the starting point:

Example 3:

$ \sum_{\color{red}{i=7}} ^{\color{red}{10}}{\color{#1299e7}{\frac{1}{i}}} =?$

Solution: $ \sum_{i=7} ^{10}{\frac{1}{i}} =\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}$

Now it is your turn!
Solve the following summation symbol examples, find the sum (you can check solutions at the bottom of this page):

Summation of a constant

In previous examples, you may have noticed that the lower summation index i also appears on the right side of the summation sign. Let's look at an example when there is only a constant next to a summation sign:

$ \sum_{\color{red}{i=1}} ^{\color{red}{2}}{\color{#1299e7}{4}} = 4+4$

In such case, we sum the constant with itself. How many times? As many times as there are numbers from the lower to the upper bound — in the example above, 2 times ( we say that the index i goes from 1 to 2).

Let's solve more similar examples:

Example 1:

$ \sum_{\color{red}{i=1}} ^{\color{red}{3}}{\color{#1299e7}{2}} = ?$

Solution: because i goes from 1 to 3, we sum 2 three times: $ \sum_{i=1} ^{3}{{2}} =2+2+2=3\cdot 2=6$

Example 2:

$ \sum_{\color{red}{i=0}} ^{\color{red}{3}}{\color{#1299e7}{2}} =?$

Solution: because i goes from 0 to 3: 0,1,2,3 - we sum 2 four times: $ \sum_{i=0} ^{3}{{2}} =2+2+2+2=4\cdot 2=8$

Solution: because i goes from 0 to 3: 0,1,2,3 - we sum 2 four times: $ \sum_{i=0} ^{3}{{2}} =2+2+2+2=\\=4\cdot 2=8$

Example 3:

$ \sum_{\color{red}{i=4}} ^{\color{red}{8}}{\color{#1299e7}{1}} = ?$

Solution: i goes from 4 to 8: 4,5,6,7,8 so there are 5 numbers. Therfore we sum 1 five times: $ \sum_{i=4} ^{8}{{1}} =1+1+1+1+1=5\cdot 1=5$

Solution: i goes from 4 to 8: 4,5,6,7,8 so there are 5 numbers. Therfore we sum 1 five times: $ \sum_{i=4} ^{8}{{1}} =1+1+1+1+1=\\=5\cdot 1=5$

So, how many times do we sum the constant?

Subtract the lower from the upper bound and add one. In the example above: $8-4+1=5$. So 5 times.
Why do we add one? Because we subtracted the lower bound, but it should be included.

Now, a couple examples for you to solve, find the sum: (you can check solutions at the bottom of this page):

Solutions

The sums are: