---
title: "Advanced Algebra: Sequences and Sums"
---

# Advanced Algebra: Sequences and Sums

## Quick Cheatsheet

### Secret-Shortcut Sum Formulas

$$
\begin{align*}
\sum_{i=1}^n i \; &= \frac{n(n+1)}{2} \\
\sum_{i=1}^n i^2 &= \frac{n(n+1)(2n+1)}{6} \\
\sum_{i=1}^n i^3 &= \frac{n^2(n+1)^2}{4}
\end{align*}
$$

### General Formulas for Sequences

* Additive:
$$
a_n = a_1 + d(n-1)
$$
* Example:
$$
-1, 3, 7, 11, 15, \ldots \implies a_n = -1 + 4(n-1) = -5 + 4n
$$
* Multiplicative:
$$
a_n = a_1m^{n-1}
$$
* Example:
$$
\frac{3}{7}, \frac{9}{14}, \frac{27}{28}, \frac{81}{56}, \ldots \implies a_n = \frac{3}{7}\left(\frac{3}{2}\right)^{n-1}
$$
* Second-Order
$$
a_n = \alpha n^2 + \beta\text{ or }a_n = \alpha n^3 + \beta
$$
* Example
$$
-1, 5, 15, 29, \ldots \implies a_n = 2n^2 - 3
$$

### Simplifying Sums

* Addition and multiplication
$$
\sum_{i=1}^n ai + b = n\cdot b + a\sum_{i=1}^ni
$$

* Weird bounds

$$
\sum_{i=5}^{17}i = \sum_{i=1}^{13}i+4 = 4(13) + \sum_{i=1}^{13}i
$$

## Sequences I: Additive Sequences

In an **additive** sequence, we start with some number $a_1$, and then to get the next term in the sequence we just add some number $d$ to the previous term[^first]. For example, if $a_1 = 5$ and $d = 7$, this would produce the sequence

$$
5, 12, 19, 26, 33, \ldots
$$

and there are a couple of different ways we can come up with a general formula for the $n$th number in the sequence.

[^first]: Here we use $d$ specifically, to stand for "difference", since it tells us the gap, or difference, between each number in the sequence

### Method 1: $n+1$ term

The way I learned it was, you start with the first number, $a_1$, then find the difference $d$, to get an equation

$$
a_n = a_1 + d(n-1)
$$

The $(n-1)$ looks weird at first, but it's just there to make sure that the **first** number in the sequence, the number we get when we plug in $n = 1$, is just exactly $a_1$. So for example, if we're given the sequence

$$
-9, -7.5, -6, -4.5, \ldots
$$

we notice that $a_1 = -9$, and that $d = 1.5$, so that the formula generating this sequence is

$$
a_n = -9 + 1.5(n-1).
$$

### Method 2: $n$ term

The way you did it in today's session was only slightly different, but works just as well in most cases. You thought about, for a given sequence, how we could write its equation with just an $n$ term instead of that uglier $n-1$ term. And to do this, you looked at the sequence *backwards* to figure out what the *previous* term would be, the term before $a_1$, so I'll call it $a_0$. In the above example, since $a_1 = -9$, you would find that $a_0 = -10.5$, and then write a similar formula but using $n$ instead of $n-1$:
$$
a_n = a_0 + dn,
$$
and so in this specific case,
$$
a_n = -10.5 + 1.5n,
$$
which works exactly just as well in this case.

### Which to Use?

The reason I'm "steering" you towards the first method, though, is just because sometimes there *won't* be an $a_0$ that you can backwards-extrapolate, and they might try to trick you with that. For example, if the sequence was
$$
\frac{5}{1}, \frac{5}{2}, \frac{5}{3}, \frac{5}{4}, \ldots
$$
then sadly the "previous" term would not be defined, since we'd find $a_0 = \frac{5}{0}$, but dividing by zero is undefined. You'll be fine, since in these cases you can just eyeball it and see, oh, the general formula here is
$$
a_n = \frac{5}{n},
$$
but mean teachers might take off points if you talk about $a_0$ or a "previous term" in these cases.

And, the nice thing is, if you really want (or your teacher asks for) a just-$n$ term, you can always do the first method and then multiply stuff out. For example, above we got $a_n = -9 + 1.5(n-1)$ using the first method, and $a_n = -10.5 + 1.5n$ using the second. But,
$$
a_n = -9 + 1.5(n-1) = -9 + 1.5n - 1.5 = -10.5 + 1.5n,
$$
so by multiplying stuff out we've obtained the second-method formula from the first!

## Sequences II: Multiplicative Sequences

So in the previous case, we always got some formula that looks like

$$
a_n = a_1 + d(n-1).
$$

We can read this out as basically saying, start at $a_1$ for the first term, then just add $d$ to get the next term in the sequence, for every term after the first term.

Multiplicative sequences have the same "structure" as additive sequences, and the only difference is that successive terms in the sequences are obtained by **multiplying** the previous term by some number, rather than adding some number to it. For example if the sequence is

$$
3, 6, 12, 24, 48, \ldots
$$

then we can multiply any term in the sequence by $2$ to get the next term. So, whereas the additive sequence formulas had the "start at $a_1$ then add $d$ to get subsequent terms" structure, formulas for multiplicative sequences will be of the form "start at $a_1$, then **multiply** $m$ to get subsequent terms". If we take the first term $a_1$, then figure out the **multiplier** $m$, we get a similar formula but where addition has been replaced by *multiplication* and multiplication has been replaced by *exponentiation*:

$$
a_n = a_1m^{n-1}
$$

Just like how the $n-1$ in the additive sequence formulas ensured that we didn't start adding until the second term onwards, the $n-1$ here ensures that we don't start multiplying until the second term onwards: if we plug in $n=1$, we get

$$
a_1m^{1-1} = a_1m^0 = a_1(1) = a_1,
$$

so that we can just toss in the first term $a_1$ here, rather than worrying about extrapolating an $a_0$ term.

### Example Problem

For a quick example problem, let's take the sequence we started with above:

$$
3, 6, 12, 24, 48, \ldots
$$

Here we see that $a_1 = 3$, and that the multiplier is $m = 2$, so we can instantly derive the general formula!

$$
a_n = 3(2)^{n-1},
$$

and like I mentioned above, if we want a cleaner formula with just $n$ in the exponent, we can use our exponent rules to simplify:

$$
a_n = 3(2)^{n-1} = 3\frac{2^n}{2^1} = \frac{3}{2}2^n,
$$

which makes sense since if we ran the sequence backwards, we'd have to divide the first term $a_1$ by $2$, which would give us the "previous term" $a_0 = \frac{3}{2}$, and our multiplier would still be $m = 2$.

## Sequences III: Second-Order Sequences

These are the weird ones, where you're trying to figure out the gap $d$, or the multiplier $m$, and you realize it's actually changing every term 😨.

For these cases, I could write a bunch of stuff, but it boils down to: the general formulas for second-order sequences are going to involve squaring $n$, cubing $n$, etc... If the gap between terms goes up by $2$ each time, we'll need an $n^2$ term. If the gap goes up by $3$ each time, we'll need an $n^3$ term, and so on. To me these are hard to memorize as "rules", so I just usually write out what these look like, and see how I need to change them to match the sequence they want me to find a formula for.

### Example Problem

If our sequence is

$$
8, 11, 16, 23, 32, 43, 56, \ldots
$$

we can try just looking at the sequence $a_n = n^2$, to start with:

$$
a_1 = 1, a_2 = 4, a_3 = 9, a_4 = 16
$$

And we see that, it's halfway there: the gaps in $1, 4, 9, 16, \ldots$ increase by $2$ each time, just like the gaps in $8, 11, 16, 23, \ldots$ do. So, all that we have to do is **shift** this $n^2$ sequence to match up with the $8, 11, 16, 23, \ldots$ terms! Since the first term, $8$, is $7$ higher than the first term of $a_n = n^2$, it looks like we just need to add $7$ to each term. And indeed, if you write it out, this gives us the formula for our weirder sequence, with just that small modification of $a_n = n^2$:

$$
a_n = n^2 + 7,
$$

since

$$
\begin{align*}
a_1 &= 1^2 + 7 = 8 \; ✅ \\
a_2 &= 2^2 + 7 = 11 \; ✅ \\
a_3 &= 3^2 + 7 = 16 \; ✅ \\
& \; \; \vdots
\end{align*}
$$

## Sums: Notation

When we write

$$
\sum_{i=1}^n f(i)
$$

this means, literally, "Take $f(i)$ and plug in $1$ for $i$, then $2$ for $i$, then $3$ for $i$, and so on up to plugging in $n$ for $i$". Meaning, any time you see this notation, it's just shorthand for a longer, written-out addition of a bunch of terms:

$$
\sum_{i=1}^n f(i) = f(1) + f(2) + \cdots + f(n).
$$

The $f(i)$ notation might be confusing, so let's just consider the most basic possibility, $f(i) = i$. This sum should look familiar to you:

$$
\sum_{i=1}^n f(i) = \sum_{i=1}^n i = 1 + 2 + \cdots + n
$$

and it's one of the sums that has a simple formula in our quick cheatsheet above.

## Simplifying: Sums *Multiplied By* a Constant

So the idea is, since we have these secret-shortcut formulas for $\sum_{i=1}^ni$, $\sum_{i=1}^ni^2$, and $\sum_{i=1}^ni^3$, if they give us some complicated-looking sum we try our best to *reduce* it down to one of these three sums, then we don't have to actually sum anything up (just plug $n$ into one of the formulas).

Our first simplifying rule is

$$
\sum_{i=1}^naf(i) = a\sum_{i=1}^nf(i),
$$

where $a$ is just any number other than $i$, like $3$ or $14$ or $\pi$. To understand this, so that you don't have to memorize it, you can always write out the sum (we'll use $f(i) = i$ like we did above):

$$
\sum_{i=1}^nai = a \cdot 1 + a \cdot 2 + \cdots + a \cdot n,
$$

and since every term there has an $a$ in it, we can pull out this common term to get

$$
a \cdot 1 + a \cdot 2 + \cdots + a \cdot n = a(1 + 2 + \cdots + n).
$$

But $1 + 2 + \cdots + n$ is just $\sum_{i=1}^ni$, which is one of our secret-shortcut sums!

$$
a(1 + 2 + \cdots + n) = a\sum_{i=1}^ni
$$

Putting it all together, then, we have:

$$
\sum_{i=1}^nai = a\sum_{i=1}^ni = a \cdot \frac{n(n+1)}{2}.
$$

## Simplifying: Sums with *Added* Constants

The other case where you can easily simplify is when your sum looks like

$$
\sum_{i=1}^n f(i) + b,
$$

where just like $a$ above, $b$ is any number besides $i$, like $3$, $14$, or $\pi$. If we write this out (so that we don't have to memorize), using $f(i) = i^2$ since I'm getting bored of $f(i) = i$, we get

$$
\sum_{i=1}^n f(i) + b = \sum_{i=1}^n i^2 + b = 1^2 + b + 2^2 + b + \cdots + n^2 + b.
$$

But, if you stare at that for a bit, you'll notice that there are $n$ terms total, and we're just adding $b$ in each term, which means that $b$ will get added $n$ times, which is exactly the definition of $n \cdot b$. Hence the name, $n$ *times* $b$. Therefore, we can pull the $b$ out of the inside of the sum, and just add an $n \cdot b$ term *outside of* the sum:

$$
\sum_{i=1}^n i^2 + b = n\cdot b + \sum_{i=1}^n i^2.
$$

And since this is one of our secret-shortcut sums, we can now get rid of the summation entirely:

$$
n\cdot b + \sum_{i=1}^n i^2 = n \cdot b + \left(\frac{n(n+1)(2n+1)}{6}\right).
$$

## Simplifying: Shifting the Bounds

One last simplifying trick: if they give you a sum with weird bounds, like

$$
\sum_{i=53}^{87}i,
$$

they're trying to trick you to see if you'll blindly apply the secret-shortcut formulas, which won't work in these cases since these formulas *require* that the sum starts with $i=1$. But, in these cases, try writing out the sum to see how we can *turn it into* a sum of *something* starting at $1$:

$$
\sum_{i=53}^{87}i = 53 + 54 + \cdots + 86 + 87
$$

Here we see that, if we started the sum at $i = 1$, but **added 52 to each term**, we'd get this exact same sum! (Being careful to figure out what our $n$ will be, by looking at the final term in the original sum and seeing what value is 52 less than it):

$$
\begin{align*}
\sum_{i=1}^{35} i + 52 &= (1+52) + (2+52) + \cdots + (34+52) + (35 + 52) \\
&= 53 + 54 + \cdots + 86 + 87,
\end{align*}
$$

and *voilà*, we have our original sum back, but with the bounds starting at $i = 1$ like we wanted! Now, notice that we've added this $+ 52$ term into the mix, but we know how to pull this out of the summation from our simplifying tools in the previous section:

$$
\sum_{i=1}^{35} i + 52 = 35(52) + \sum_{i=1}^{35}i,
$$

and we're finally at one of our secret-shortcut formulas, so we apply that to get a final whole-number answer without any summations at all:

$$
35(52) + \sum_{i=1}^{35}i = 35(52) + \frac{(35)(36)}{2} = 35\left(52 + \frac{36}{2}\right) = 35(70) = 2450
$$

(which you can check using [WolframAlpha](https://www.wolframalpha.com/input?i=sum+of+x+from+53+to+87))