Core Pure - Series I

If you’ve got to this point, you must be familiar with sums. It’s fine writing short sums explicitly, and sometimes allow the reader to guess what terms are missing. If you write 1 + 2 + 3 + \ldots + 100, there is likely no confusion that you mean the sum of all integers from 1 to 100.

However, there are some cases where the patterns are not that clear. If we want to make very precise statements about sums (which we often want to in maths), we can find this notation to be unsophisticated and unwieldly. That is why we employ “sigma” (\Sigma) notation for sums. In short, if we have a list of n numbers, a_1, a_2, \ldots, a_n, we write their sum as:

\displaystyle \sum_{i = 1}^n a_i = a_1 + a_2 + \ldots + a_n

we call each a_1, a_2, \ldots, a_n terms of the sum \sum_{i = 1}^n a_i.

You can use any letter instead of i. For example, it would not matter if you wrote \displaystyle \sum_{j = 1}^n a_j. Some people emphasise the fact that it doesn’t matter exactly what symbol is put here by calling it a “dummy variable”. I, however, think this is unhelpful. It would be bad practice though, to re-use any symbols that are already defined. We would not like to write \displaystyle \sum_{n = 1}^n a_n for instance, so watch out for that.

We can work out some “easy” results. We know that since c(a + b) = ac + bc, that we can write:

\displaystyle \sum_{i = 1}^n ca_i = c \sum_{i = 1}^n a_i

You can also see that sums “split up” quite nicely. If we have two lists a_1, a_2, \ldots, a_n and b_1, b_2, \ldots, b_n:

\displaystyle \sum_{i = 1}^n (a_i + b_i) = \sum_{i = 1}^n a_i + \sum_{i = 1}^n b_i

If you write out the first few terms of each of these sums, you’ll see that all we’re changing is the order in which we’re adding up the terms.

One last trick I’ll talk about is sums that don’t start from 1. Sometimes we might know what:

\displaystyle \sum_{i = 1}^n a_i

is for each n, but we might want to find, say:

\displaystyle \sum_{i = k + 1}^n a_i

for k < n. In other words, we only want to sum the terms of \sum_{i = 1}^n a_i beyond the k th term. Well, we know that the sum of the first k terms is:

\displaystyle \sum_{i = 1}^k a_i

These are essentially the terms from \sum_{i = 1}^n a_i that we don’t want - so we simply take this away from \sum_{i = 1}^n a_i to get:

\displaystyle \sum_{i = k + 1}^n a_i = \sum_{i = 1}^n a_i - \sum_{i = 1}^k a_i

Now we’ll work out a few common sums. We can see that:

\displaystyle \sum_{i = 1}^n 1 = \underbrace{1 + 1 + \ldots + 1}_{n \text { times}} = n \times 1 = n

We will now look at:

\displaystyle \sum_{i = 1}^n i^p

for p = 1, p = 2, p = 3.

The first sum has a very nice derivation. We don’t have the tools for “nice” derivations of the other two, so we will settle for (very robotic and boring) proofs by induction rather than any interesting tricks.

The method we will use for the sum \sum_{i = 1}^n i was supposedly thought of by legendary mathematician Carl Friedrich Gauss when he was in primary school, and his teacher asked him to sum the first 100 natural numbers, and Gauss figured it out in minutes. The trick is to write:

\begin{align*}\sum_{i = 1}^n i & = 1 + 2 + 3 + \ldots + n \\ & = n + (n - 1) + (n - 2) + \ldots + 1\end{align*}

where we’ve simply wrote the sum “backwards” in the second instance. Notice that if we now add these two representations we have:

\begin{align*}2 \sum_{i = 1}^n i & = (n + 1) + (n - 1 + 2) + (n - 2 + 3) + \ldots + (n + 1) \\ & = \underbrace{(n + 1) + (n + 1) + (n + 1) + \ldots + (n + 1)}_{n \text { times}} \\ & = n \times (n + 1)\end{align*}

And we finish off dividing by two giving:

\boxed{\displaystyle \sum_{i = 1}^n i = \frac {n(n + 1)} 2}

So we actually find:

1 + 2 + 3 + \ldots + 100 = \dfrac {100 \times 101} 2 = 50 \times 101 = 5050

This is a very simple and elegant proof that looks like it was plucked from nowhere. You’ll see a lot of these in maths, but with practice you’ll start spotting patterns that would have baffled you a few years ago.

We have:

\boxed{\displaystyle \sum_{i = 1}^n i^2 = \frac 1 6 n(n + 1)(2n + 1)}

Note that if:

\displaystyle \sum_{i = 1}^n i^2 = \frac 1 6 n(n + 1)(2n + 1)

then we have:

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

here we’ve just factored out \dfrac 1 6 (n + 1). We can write:

\begin{align*}\frac 1 6 (n + 1) (n(2 n + 1) + 6(n + 1)) & = \frac 1 6 (n + 1) (2n^2 + n + 6n + 6) \\ & = \frac 1 6 (n + 1)(2n^2 + 7n + 6) \\ & = \frac 1 6 (n + 1)(n + 2)(2n + 3) \\ & = \frac 1 6 (n + 1)(n + 2)(2(n + 1) + 1)\end{align*}

This is our inductive step. See if you can fill in the rest of the details of a proof.

Similarly, we can prove that:

\boxed{\displaystyle \sum_{i = 1}^n i^3 = \frac {n^2 (n + 1)^2} 4}

The eagle-eyed reader might notice that we therefore have:

\displaystyle \sum_{i = 1}^n i^3 = \left(\sum_{i = 1}^n i\right)^2

This has a nice geometric significance as we’ll look at in the exercises. You may have even suspected that:

\displaystyle \left(\sum_{i = 1}^n i\right)^2 = \sum_{i = 1}^n i^2

but this would imply that a^2 + b^2 = (a + b)^2 (!!!)

Try proving this by induction.

Exercises:

(i) [A2] Noting that (n + 1)^3 - n^3 = 3n^2 + 3n + 1, use the method of differences to prove that:

\displaystyle \sum_{i = 1}^n i^2 = \frac 1 6 n (n + 1)(2n + 1)

(ii) Say you have a bunch of cubes of volume 1 unit. With these, you make n cubes of side length 1, 2, 3, \ldots n. What is the sum of the volumes of these cubes? By rearranging the smaller cubes into a cuboid of height 1, and finding said cuboids volume, show that:

\displaystyle \left(\sum_{i = 1}^n i\right)^2 = \sum_{i = 1}^n i^3

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The exercises are great!

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