Sum of Natural Numbers

Series of Positive Terms

By now you have learned about series as sums, finite or infinite, of terms of sequences. The series you will learn here are series of positive terms, that is, the sum of the terms of sequences formed exclusively by positive terms,

\[ \sum_{r}^{}a_r = a_1+a_2+a_3+..., \]

where \(a_r\) is positive for every \(r\).

Throughout this article, we will provide you with tools to find the exact value of series of positive terms.

Adding up many positive numbers properties

A series of positive terms is an elegant mathematical term for the sum of a sequence of some or many positive numbers.

Try to add up the following, \[ 1+1+1+1+1+1+1+1. \]

Easy, right? What about this,

\[ \sum_{r=1}^{8}1 .\]

It’s the same! The only difference is that it is written with sigma notation.

\[ \sum_{r=1}^{8}1=1+1+1+1+1+1+1+1=8 .\]

You can generalize this sum a little more by doing,

\[ \sum_{r=1}^{n}1=\underset{n\text{ times}}{\underbrace{1+1+1+\cdots+1}}=n. \]

This is the first sum we want you to memorize. It will be very useful for you to solve other sums. Let's apply it to an example.

The last method of resolution is the application of a rule you may already know:

Constant Multiple Rule. \(\sum_{r=1}^{n}kf(r)=k\sum_{r=1}^{n}f(r) \), where \(k\) is a constant and \(f\) is a function of \(r\).

In the case of the example, after turning \(3\) into \(3\times 1\), \(k=3\) and \(f(r)=1\).

To recall the other rules of sigma notation, please return to our article on Series.

Sum of the First 100 Natural Numbers

Now let's tackle this very classic sum

\[ 1+2+3+4+5+\cdots+99+100\,. \]

How much do you think this sum is? How are you going to approach it? First thing, you can write this sum with sigma notation

\[ \sum_{r=1}^{100}r\,. \]

Will the rules of sigma notation work here? It seems they won't serve us, and they don’t.

Next attempt: if you can find an algebraic expression for the sequence

\[ 1,2,3,4,5,\cdots,99,100,\]

maybe you could work it out later.

Let's look at three approaches, all of them very interesting, to this challenge.

Historical approach

This approach does not involve finding an algebraic expression for the sequence of the first 100 natural numbers, but a pattern that is present in them.

The story goes that one day a mathematics teacher asks the students in a class to calculate the sum of all the whole numbers from 1 to 100. The students picked up their slates and began working on the challenge.

Not before long, a student gets up and places his slate on the teacher's desk. After a while longer, the teacher's desk is filled with slates with the students' resolutions.

Now it was the teacher's turn to check each student's solution. Wrong answer after wrong answer, he arrived at the last slate, that of the first student to solve the challenge in no time. It was correct, and it was the answer of the young Carl Friedrich Gauss, who would become a great mathematician.

Here is what he thought. Gauss observed that adding 1 to 100 gave 101, and 2 to 99 also gave 101, as did 3 to 98. Then he noticed that there were 50 pairs of numbers between 1 and 100, included, which added up to 101.

\[ \underset{1+100=101}{\underbrace{1,\overset{2+99=101}{\overbrace{2,\underset{3+98=101}{\underbrace{3,\cdots,\overset{49+52=101}{\overbrace{49,\underset{50+51=101}{\underbrace{50,51}},52}},\cdots,98}},99}},100}} \]

Finally, he multiplied the sum \(101\) by the \(50\) pairs, resulting in \(5050\), which was the answer to the challenge.

Classical approach

This is the usual approach taken in school and involves noticing that the sequence

\[ 1,2,3,4,\cdots, 99,100,\]

is an arithmetic sequence of common difference 1.

Recalling that the general term of an arithmetic sequence is

\[ a_n=a_1+(n-1)d, \]

where \(a_n\) is the \(n^{\text{th}}\) term in the sequence, \(a_1\) is the 1st term and \(d\) is the common difference, then for the previous sequence you'll have \( a_1=1, \, n=100, \, d=1 \) and

\[ a_n=1+(n-1)\times1=n. \]

Now, it is time to remember that the sum of the first \(n\) terms of an arithmetic sequence is

\[ S_n={n \over 2}(a_1+a_n)={n \over 2}(1+n). \]

In the sum you want to calculate, \(n=100\) and \(a_n=a_{100}=100\). So, by inputting this in the last formula you have

\[ S_{100}={100 \over 2}(1+100)=50\times 101=5050. \]

Can you see how this completely algebraic method coincides with the reasoning of the young Carl Gauss? You even ended up with the same numbers – \(50\) which is the pair of numbers that sum \(101\). Interesting, right?

Out-of-the-box approach

Let's represent numbers with dots. So, the number 1 will be represented with one dot •, the number 2 with two dots • •, and so on. With this in mind, we present to you the triangular numbers, figure 1.

Fig. 1 - Triangular numbers.

The triangular numbers \(1, 3, 6, 10, ...\), form a sequence that we will denote with \((t_n)\). This means that \(t_1=1,t_2=3,t_3=6\), and so on.

Now, notice that each triangular number is equal to the sum of the first \(n\) natural numbers. For instance,

\[ \begin{align} t_1&=1,\\ t_2&=3=1+2, \\ t_3&=6=1+2+3, \\ t_4&=10=1+2+3+4. \end{align} \]

This means then that \[t_n=1+2+3+\cdots+n-1+n,\]

in particular, for \(n=100\), \[t_{100}=1+2+3+\cdots+99+100\, . \]

So, if you can find an expression for the general term \(t_n\), you'll also get a formula for the sum of the first \(n\) natural numbers and, in particular, the sum of the first \(100\) natural numbers. And to achieve this, let's explore some interesting relationships between two consecutive triangular numbers.

The difference between two consecutive triangular numbers is n

Here, you will explore the pattern that allows you to get the next triangular number from the previous one.

Experimenting with the triangular numbers from above, you have,

• for \(n=2\), \( t_2=3\) which is the same as \(t_1+2\);

• for \(n=3\), \( t_3=6\) which is the same as \(t_2+3\);

• for \(n=4\), \( t_4=10\) which is the same as \(t_3+4\).

This leads you to the conclusion that \[t_n=t_{n-1}+n,\] which rearranging, you get \[t_n-t_{n-1}=n, \]

for all natural number \(n\).

The sum of two consecutive triangular numbers is n²

Let's now 'transform' triangles into squares. We are going to turn triangular numbers into square numbers.

For simplicity, let's look at the graphical representation of the triangular number 6, \(t_3\), and let's add dots to that representation until we get a square, figure 2.

Figure 2 - Two consecutive triangular numbers form a square number

Notice that to get a square, we added to the triangular number 6, \(t_3\), the triangular number 3, \(t_2\). If you try the same thing with the triangular number 10, you will see that to get a square, you add to 10, \(t_4\) the triangular number 6, \(t_3\). This means that the sum of two consecutive triangular numbers is equal to a square number, that is

\[ t_n+t_{n-1}=n^2\, . \]

The sum from 1 to 100

Now we apply a little algebraic calculation.

You know that \(t_n-t_{n-1}=n\), that by reorganizing you also have

\[t_{n-1}=t_n-n\, .\]

And you will apply this to the previous formula,

\[ \begin{align} t_n+t_{n-1}&=n^2 \\ t_n+(t_n-n)&=n^2 \\ t_n+t_n&=n^2 +n \\ 2t_n&=n(n+1) \\ t_n&={n \over 2}(1+n)\, . \end{align} \]

Now, replacing \(n\) by \(100\) you get

\[ t_{100}={100 \over 2}(1+100)=50\times 101=5050\, . \]

Sum of the First n Natural Numbers: Formula

By now, you must have understood that the formula that lets you calculate the sum of the first \(n\) natural numbers is

\[ \sum_{r=1}^{n}r={n \over 2}(n+1)\, . \]

And this is the second formula we want you to memorize because it will help you solve other series. Let's see that in the next examples.

In the second example, you saw a sum that does not start at \(1\), but at \(15\). Whenever you want to know the sum of a list of natural numbers that starts at a number other than \(1,\) let's say \(k\), what you should do is consider that same list as if it started at \(1\) and then subtract the excess, from \(1\) to \(k-1\). We summarize this in the following rule,

Rule. \(\sum_{r=k}^{n}f(r)=\sum_{r=1}^{n}f(r)-\sum_{r=1}^{k-1}f(r)\), where \(k\) and \(n\) are natural numbers, \(f\) is a function of \(r\).

Proof of the Sum of the First n Natural Numbers

Earlier, you saw three approaches to the sum of the first \(100\) natural numbers. In two of these approaches, you saw the formula for the sum of the first \(n\) natural numbers. Here we will present you with a formal demonstration of that formula.

To simplify notation, here we will call the sum of \(1\) through \(n\), \(S_n\), meaning \(S_n= \sum_{r=1}^{n}r\).

We begin by writing the sum of the first \(n\) natural numbers from \(1\) to \(n\) in two ways,

\[ \begin{align} S_n&=1+2+\cdots+n-1+n \\ S_n&=n+n-1+\cdots+2+1 \end{align} \]

Next, we sum these two formulas,

\[ \begin{align} S_n+S_n&=(1+n)+(2+n-1)+\cdots+(n-1+2)+(n+1) \\ 2S_n&=(n+1)+(n+1)+\cdots+(n+1)+(n+1) \end{align} \]

Now, we have to count how many \((n+1)\) we have, and there are \(n\) of them,

\[ \begin{align} 2S_n&=(n+1)+(n+1)+\cdots+(n+1)+(n+1)=n(n+1) \\ S_n&={n \over 2}(n+1).\end{align} \]

Sum of the First n Even Natural Numbers

You now come to the section where you will learn the third important formula that you want to memorize, and it has to do with the sum of the first \(n\) even natural numbers. We have the advantage that we already have the formula for the sum of the first \(n\) naturals, as it will be needed here.

So, the sum you want to determine is

\[ 2+4+6+8+10+\cdots\, ,\]

but for the first \(n\) even numbers.

The sequence of even numbers has the expression \((2n)\), where \(n\) is a natural number. You can then complete the sum of the first \(n\) even numbers as

\[ 2+4+6+8+10+\cdots+2n\text{ or }\sum_{r=1}^{n}2r\]

Now, applying the Constant Multiple Rule of the sigma notation you have

\[ \sum_{r=1}^{n}2r=2\sum_{r=1}^{n}r=2\times {n \over 2}(n+1)=n(n+1)\, .\]

Let's see some examples of the application of this formula.

Sum of the First n Odd Natural Numbers

You now come to the section where you will learn the fourth important formula that you want to memorize, and it has to do with the sum of the first \(n\) odd natural numbers. This formula is easily deduced from the application of known sums and a few other rules.

Knowing that \((2n-1)\) is the sequence of odd numbers, where \(2n-1\) is the \(n^{\text{th}} \) odd number, then

\[ 1+3+5+7+9+\cdots+2n-1=\sum_{r=1}^{n}(2r-1) .\]

Applying the Difference Rule of the sigma notation, you have\[ \sum_{r=1}^{n}(2r-1)=\sum_{r=1}^{n}2r-\sum_{r=1}^{n}1.\]

Now, apply the formula of the sum of the first \(n\) even numbers and the sum of \(n\) 1's

\[ \begin{align} \sum_{r=1}^{n}2r-\sum_{r=1}^{n}1&=n(n+1)-n \\ &=n((n+1)-1) \\&=n(n)=n^2. \end{align} \]

Let's see some examples of the application of this formula.

Examples of Sums of Natural Numbers

Now that you have all the tools we wanted you to learn here, let's get to some more application examples of this topic.

The first example serves as a warm-up.

The second example requires all your wit on this topic.