IB Mathematics Applications & Interpretation SL 1.2—sequences and Sigma Notation Notes

Why Sequences And Series Matter

1. Many real situations produce an ordered list of numbers: the money in a savings account after each monthly deposit, the number of seats in each row of a theatre, or a salary that rises by a set raise each year.
2. Once we spot the pattern behind such a list, we can jump straight to any entry, or total a long run of them, without writing every number out by hand.

What Makes A Sequence Arithmetic

1. A sequence is an ordered list of numbers, and each number in it is called a term.
2. We label the terms $u_1, u_2, u_3, \dots$, where the small number underneath gives the term's position in the list.
   1. $u_1$ is the first term,
   2. $u_2$ the second,
   3. $u_n$ means the term in position $n$.
3. A sequence is arithmetic when you get from each term to the next by adding the same fixed number every time.
4. That fixed number is the common difference $d$, and you find it by subtracting any term from the one straight after it.
5. In symbols, $d = u_2 - u_1 = u_3 - u_2$, and so on.
6. If that difference is not the same for every pair of neighbours, the sequence is not arithmetic.

A Formula For The nth Term

1. To reach a term far along the list, adding $d$ over and over is slow, so we want a direct formula.
2. Look at how the terms are built from the first one: $u_1$, then $u_1 + d$, then $u_1 + 2d$, then $u_1 + 3d$.
3. The number of $d$'s added is always one less than the position, so to reach position $n$ you add $d$ exactly $(n-1)$ times. $$ u_n = u_1 + (n-1)d $$
   1. $u_n$ is the term you want,
   2. $u_1$ the first term,
   3. $d$ the common difference,
   4. $n$ the position.
4. The same formula runs backwards: setting $u_n$ equal to a known value and solving for $n$ tells you which position a term sits in.

Adding The Terms: The Arithmetic Series

1. A series is what you get when you add up the terms of a sequence, and $S_n$ means the sum of the first $n$ terms.
2. There is a shortcut, and the trick that reveals it is to pair terms from the two ends inwards.
3. Write the sum forwards, then write it again backwards underneath, and add the two lines term by term. $$ 2S_n = \underbrace{(u_1 + u_n) + (u_1 + u_n) + \dots + (u_1 + u_n)}_{n \text{ pairs}} = n(u_1 + u_n) $$
4. Every vertical pair adds to the same total $u_1 + u_n$, and there are $n$ of them, so halving gives the sum. $$ S_n = \frac{n}{2}(u_1 + u_n) = \frac{n}{2}\big(2u_1 + (n-1)d\big) $$
5. The second form comes from replacing $u_n$ with $u_1 + (n-1)d$, and it is the one to use when you do not yet know the last term.

Sigma Notation: A Compact Way To Write A Sum

1. When a sum has many terms, writing them all out is slow, so we use the Greek capital letter sigma, $\sum$, as a shorthand for "add these up".
2. The notation has three parts around the $\sum$, and you can read them like a recipe.
   1. The letter written below, here $k$, is the counter, and the value beside it is where the counter starts, for example $k = 1$.
   2. The number written above is the last value the counter reaches, for example $n$.
   3. The expression to the right is the rule that turns each counter value into one term of the sum.
3. To work it out, substitute each counter value in turn, one term at a time, and add all the results. $$ \sum_{k=1}^{4}(2k+1) = (2\cdot 1+1) + (2\cdot 2+1) + (2\cdot 3+1) + (2\cdot 4+1) =$$ $$= 3 + 5 + 7 + 9 = 24 $$
4. Read aloud, that says "add up $2k+1$ as $k$ runs from 1 to 4".
5. For an arithmetic series the rule for each term is just the $n$th-term formula, so the whole sum can be written compactly. $$ S_n = \sum_{k=1}^{n} \big(u_1 + (k-1)d\big) $$

Where Arithmetic Sequences Show Up

1. Arithmetic models fit anything that changes by a fixed amount each step, such as simple interest added each year, seats gained per row in a theatre, or a salary rising by a set raise.
2. Real data is rarely perfectly arithmetic, so you may have to approximate the common difference before using the model to predict.