Two short pieces of support, for the two things this module assumes and does not teach. Both are worked on the module's own numbers, so nothing here is a detour — you are doing the Bernoulli unit's arithmetic in advance, slowly.
Read either, both or neither. Nothing depends on them and they are outside the timing budget. If the Bernoulli unit went fine, you did not need them.
1. A fraction refresher, on this module's own values
The module keeps everything as exact fractions and never converts to decimals.
That is not fussiness. -1/30 is exactly -1/30; -0.0333 is not, and the
whole point of the verification lab is comparing values that either agree
exactly or do not.
Adding fractions
To add, both parts need the same bottom number (the denominator).
$$\frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$$
1/3 became 2/6 by multiplying top and bottom by 2. Multiplying top and bottom
by the same number never changes a fraction's value — it renames it.
With the module's own numbers:
$$\frac{1}{6} + \left(-\frac{1}{30}\right) = \frac{5}{30} - \frac{1}{30} = \frac{4}{30} = \frac{2}{15}$$
Two things happened at the end. 4/30 was simplified to 2/15 by dividing
top and bottom by 2. And adding a negative is subtracting — the sign travels with
the fraction, which is where most slips in this module come from.
Multiplying fractions
Multiply the tops, multiply the bottoms. No common denominator needed.
$$\frac{1}{6} \times \frac{3}{5} = \frac{3}{30} = \frac{1}{10}$$
Why the signs matter here
Bernoulli numbers alternate in sign, and the single most common error in this module is losing one. If you take nothing else from this section, take this: when you reach an equation like
$$\frac{1}{6} + 5 B_4 = 0$$
the term with B_4 in it must be the negative of everything else, because the
whole thing adds to zero. So 5 B_4 = -1/6, and B_4 = -1/30. If you get
+1/30, you have dropped the sign at exactly the step where the module expects
you to.
2. How to read a recurrence
A recurrence defines each value using the values before it. You cannot get the fifth one without the earlier ones, which is why the module has you build a table from the bottom up.
The one you will meet looks like this:
$$\sum_{j=0}^{n} \binom{n+1}{j} B_j = 0 \quad \text{for } n \ge 1$$
Three pieces of notation, one at a time.
The summation sign
$\sum$ says "add up a list". Underneath it is where the counter starts, above it is where it stops:
$$\sum_{j=0}^{3} j = 0 + 1 + 2 + 3 = 6$$
j takes each value in turn, and whatever follows the $\sum$ is written out once
per value and added together. That is all it is: a compact way of writing a long
sum without inventing names for every term.
The binomial coefficient
$\binom{n+1}{j}$ is a plain number, not a fraction — the horizontal line is missing on purpose. It is computed as
$$\binom{a}{b} = \frac{a!}{b!\,(a-b)!}$$
where a! means a × (a-1) × ... × 1, and 0! = 1 by convention. For small
numbers it is quicker to read them off Pascal's triangle:
| a | the row |
|---|---|
| 0 | 1 |
| 1 | 1, 1 |
| 2 | 1, 2, 1 |
| 3 | 1, 3, 3, 1 |
| 4 | 1, 4, 6, 4, 1 |
| 5 | 1, 5, 10, 10, 5, 1 |
$\binom{5}{2}$ is the entry at position 2 (counting from 0) in row 5, which is 10.
Putting it together
Take the recurrence at n = 3. The sum runs j = 0, 1, 2, 3, and the
coefficients come from row n+1 = 4, which is 1, 4, 6, 4:
$$\binom{4}{0} B_0 + \binom{4}{1} B_1 + \binom{4}{2} B_2 + \binom{4}{3} B_3 = 0$$
$$1 \cdot B_0 + 4 B_1 + 6 B_2 + 4 B_3 = 0$$
Substituting the values you already have — B_0 = 1, B_1 = -1/2, B_2 = 1/6,
B_3 = 0:
$$1 + 4\left(-\frac{1}{2}\right) + 6\left(\frac{1}{6}\right) + 0 = 1 - 2 + 1 = 0$$
Which checks out, and is worth doing once: it confirms you can read the notation, using only values the module has already given you.
Now n = 4, where the answer is not yet known. Coefficients from row 5:
1, 5, 10, 10, 5.
$$1 \cdot B_0 + 5 B_1 + 10 B_2 + 10 B_3 + 5 B_4 = 0$$
$$1 + 5\left(-\frac{1}{2}\right) + 10\left(\frac{1}{6}\right) + 0 + 5 B_4 = 0$$
$$1 - \frac{5}{2} + \frac{5}{3} + 5 B_4 = 0$$
The first three add to 1/6 — check that yourself, it is the fraction practice
from section 1. So 1/6 + 5 B_4 = 0, and by the sign rule above,
B_4 = -1/30.
That is the derivation the Bernoulli unit asks for, and you have just done it. Doing it there again, without this page open, is the part that matters — the point of that unit is that you build the reference rather than being handed it.
If this was still hard
That is a finding rather than a failure, and it has a specific answer: the route without the mathematics reaches the same method through a case that has no arithmetic in it at all. It is a full path, not a reduced one. The routes page lists it, and Unit 0 explains how to choose.