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Arithmetic series

This topic will cover arithmetic sequences and series by:

  1. defining an arithmetic sequence,
  2. defining an arithmetic series,
  3. calculating the sum $1 + 2 + \cdots + n$,
  4. proving $\sum_{k = 1}^n k = \frac{n(n + 1)}{2}$ using a proof by induction,
  5. showing how a proof by induction can be used to disprove a formula,
  6. summarizing this topic, and
  7. exercises.

1. Definition of an arithmetic sequence

A finite arithmetic sequence $a$ is any sequence of the form:

$a = \left(a[0], a[0] + d, a[0] + 2d, \ldots, a[0] + nd\right)$

where $a[0]$ and $d$ are given real or complex numbers. Sometimes, we will give a label to each entry in this sequence, where such a sequence is defined recursively as follows: given an initial entry $a[0]$, we define $a[k] = a[k - 1] + d$; that is, the next entry is always a fixed amount $d$ different from the previous entry in the sequence. In this case, it is reasonably easy enough to come up with an explicit representation of an arithmetic sequence:

$a[k] = a[0] + kd$

for $k = 0, 1, 2, 3, \ldots, n$.

2. Definition of an arithmetic series

Now, there are many situations in the engineering world where such sequences occur, and there are many times where we may want to compute the sum of such a sequence:

$\sum_{k = 0}^n a[k] = \sum_{k = 0}^n (a[0] + kd) = a[0] + (a[0] + d) + (a[0] + 2d) + \cdots + (a[0] + nd)$.

This is called an arithmetic series. In order to calculate such a series, we note that we may split the sum into two separate sums:

$(a[0] + a[0] + \cdots + a[0]) + (d + 2d + \cdots + nd)$.

where we are summing $a[0]$ $n + 1$ times, so this simplifies to

$(n + 1)a[0] + (d + 2d + \cdots + nd)$.

The second sum is one where we can factor out a common multiplier of $d$, so we are left with this sum being

$(n + 1)a[0] + d(1 + 2 + \cdots + n)$.

Thus, we are really only interested in knowing the value of

$1 + 2 + \cdots + n$.

This arithmetic series may also be represented by the sum

$\sum_{k = 1}^n k$,

meaning that the variable $k$ takes on all values between $1$ to $n$, and we are then summing those values together. If you wanted to calculate the sum-of-the-squares of the integers from $1$ to $10$, you would write

$\sum_{k = 1}^{10} k^2 = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385$.

3. Calculating $1 + 2 + \cdots + n$

The easiest way to find the sum of the first $n$ integers is to write down the sum twice, only the second in reverse order:

$\begin{array}{cccccccccc} 1&+&2&+&3&+ \cdots +&(n - 2)&+&(n - 1)&+&n \\ n&+&(n - 1)&+&(n - 2)&+ \cdots +&3&+&2&+&1 \end{array}$

Note that each column sums to $n + 1$, so thus, these two sums, when added together pairwise yields

$\begin{array}{ccccccccccc} \ &1&+&2&+&3&+ \cdots +&(n - 2)&+&(n - 1)&+&n \\ {\bf +}&n&+&(n - 1)&+&(n - 2)&+ \cdots +&3&+&2&+&1\\ \hline &(n + 1)&+&(n + 1)&+&(n + 1)&+ \cdots +&(n + 1)&+&(n + 1)&+&(n + 1) \end{array}$

As each column adds to $n + 1$, the total sum is nothing more than $n$ times $n + 1$, so the sum of these two series is $n(n + 1)$. As we added the original sum twice, the original sum thus half this value, or is $\frac{n(n + 1)}{2}$. Thus, remember the formula:

$\sum_{k = 1}^n k = \frac{n(n + 1)}{2}$.

If you are therefore ever asked to determine what is, for example:

$42 + 42.6 + 43.2 + 43.8 + 44.4 + \cdots + 71.4 + 72$,

you can now immediately see that this is

$\sum_{k = 0}^n (42 + 0.6k)$,

and thus, the solution is

$\sum_{k = 0}^{50} (42 + 0.6k) = 51\cdot 42 + 0.6\frac{20\cdot 21}{2}$,

and with a little arithmetic, we see that

$51\cdot 42 = 2142$ and $\frac{50\cdot 51}{2} = 1275$,

so

$\sum_{k = 0}^{50} (42 + 0.6k) = 2142 + 0.6\cdot 1275 = 2907$.

The arithmetic series $\sum_{k = 0}^n k = \frac{n(n + 1)}{2}$ is fundamental to understanding the run time and memory use of algorithms. You will see, for example, that some sorting algorithms are simply fundamentally slow in general. This information is only deducible if one understands the relationship that this sum is essentially of the from $\frac{n^2}{2} + \frac{n}{2}$.

Incidentally, given this, we now have the following general formula for the sum of an arithmetic sequence:

$\sum_{k = 0}^n \left( a[0] + kd \right ) = (n + 1)a[0] + d\frac{n(n + 1)}{2} = \frac{(2a[0] + dn)(n + 1)}{2}$,

however, this author would strongly suggest you not memorize this formula.

4. Proving $\sum_{k = 1}^n k = \frac{n(n + 1)}{2}$ by induction

We deduced the formula for the arithmetic series by summing the sum twice and adding corresponding entries. We will now prove that this formula is correct by induction.

Theorem
The arithmetic series $\sum_{k = 1}^n k = \frac{n(n + 1)}{2}$.

Proof:
To prove a statement is true by induction, we must:

  1. Prove that the statement is true for the base case, in this case, for $n = 1$.
  2. Assume the statement is true for an arbitrary value of $n$, and
  3. Under the assumption that it is true for $n$, we must show that the statement continues to be true for $n + 1$.

In this case, the statement for $n = 1$ is true:

$\sum_{k = 1}^1 k = 1$,

and

$\frac{1 \cdot 2}{2} = 1$.

As these are the same, the statement is correct when $n = 1$.

Next, we assume the statement is true for $n$, so then we consider the statement for $n + 1$: we want to show that

The arithmetic series $\sum_{k = 1}^{n + 1} k = \frac{(n + 1)((n + 1) + 1)}{2} = \frac{(n + 1)(n + 2)}{2}$.

Now,

$\sum_{k = 1}^{n + 1} k = 1 + 2 + 3 + \cdots + (n - 1) + n + (n + 1)$,

so we see that

$\sum_{k = 1}^{n + 1} k = \left(\sum_{k = 1}^n k\right) + (n + 1)$.

By assumption, the sum up to $n$ is $\frac{n(n + 1)}{2}$, so we can substitute this in:

$\sum_{k = 1}^{n + 1} k = \frac{n(n + 1)}{2} + (n + 1)$,

so finding a common denominator, we have:

$\sum_{k = 1}^{n + 1} k = \frac{n(n + 1) + 2(n + 1)}{2}$

and factoring out a common $(n + 1)$, we have:

$\sum_{k = 1}^{n + 1} k = \frac{((n + 1)(n + 2)}{2}$,

which is the statement evaluated at $n + 1$, so what we set out to prove. ▮

Now, many students only ever see a proof by induction when it works, so here is an example how a proof by induction can be used to show a formula is wrong.

5. Using a proof by induction to show a statement is wrong

Let us consider a slightly different case: suppose someone comes to you with the claim:

$\sum_{k = 1}^n k = \frac{n^3 - 3n^2 + 14n - 6}{6}$,

and that person demonstrates that it is indeed correct for $n = 1, 2$ and $3$. If you were to try to prove this by induction, we have the base case, so now we continue:

$\sum_{k = 1}^{n + 1} k = \frac{(n + 1)^3 - 3(n + 1)^2 + 14(n + 1) - 6}{6} = \frac{n^3 + 11 + 6}{6}$.

Substituting in our assumed formula into the equation

$\sum_{k = 1}^{n + 1} k = \left(\sum_{k = 1}^n k\right) + (n + 1)$

yields

$\sum_{k = 1}^{n + 1} k = \frac{n^3 - 3n^2 + 14n - 6}{6} + (n + 1)$

so finding a common denominator, we have

$\sum_{k = 1}^{n + 1} k = \frac{n^3 - 3n^2 + 14n - 6 + 6n + 6}{6} = \frac{n^3 - 3n^2 + 20n}{6}$,

and this is not equal to the formula we found above: $\frac{n^3 + 11 + 6}{6}$. Thus, because $\frac{n^3 + 11 + 6}{6} \ne \frac{n^3 - 3n^2 + 20n}{6}$, we have a contradiction.

The only assumption we made was that $\sum_{k = 1}^n k = \frac{n^3 - 3n^2 + 14n - 6}{6}$, and therefore this assumption must be wrong.

Therefore, $\sum_{k = 1}^n k \ne \frac{n^3 - 3n^2 + 14n - 6}{6}$ and the other person is wrong. ▮

6. Summary

Given the arithmetic sequence

$a[0], a[0] + d, a[0] + 2d, a[0] + 3d, \ldots, a[0] + nd$,

we can calculate this by observing that

$1 + 2 + 3 + \cdots + n = \frac{n(n + 1)}{2}$

and thus the series

$a[0] + (a[0] + d) + (a[0] + 2d) + (a[0] + 3d) + \cdots + (a[0] + nd) = (n + 1)a[0] + d\sum_{k = 1}^n k = (n + 1)a[0] + d\frac{n(n + 1)}{2}$

can be calculated using

$(n + 1)a[0] + d\sum_{k = 1}^n k = (n + 1)a[0] + d\frac{n(n + 1)}{2}$.

Now, given the sequence

$(6.273, 6.598, 6.923, 7.248, 7.573, 7.898, 8.223, 8.548, 8.873)$

we can calculate the sum of these by observing it is an arithmetic series with nine terms, $x[0] = 6.273$ and $d = 0.325$, then the sum is

$9 \cdot 6.273 + 0.325 \frac{8 \cdot 9}{2} = 56.457 + 0.325 \cdot 36 = 68.157$.

7. Exercises

1. Calculate each of the following series:

  1. $1 + 2 + 3 + 4 + 5 + 6$
  2. $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10$
  3. $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + \cdots + 100$

2. What is the sum of the first one million integers?

3. Calculate each of the following series:

  1. $(73.2, 72.3, 71.4, 70.5, 69.6, 68.7, 67.8)$
  2. $(-3.2, 60.2, 123.6, 187.0, 250.4, 313.8, 377.2, 440.6, 504.0)$
  3. $(-3.2, 60.2, 123.6, 187.0, 250.4, 313.8, 377.2, 440.6, 504.0)$

4. What is the sum of this arithmetic sequence?

$(45.9, 42.1, 38.3, 34.5, 30.7, 26.9, 23.1, \ldots, -5106.9)$