Two short pieces of support, for the two things this course assumes and does not teach. Both are worked on the course'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 course's own values
The course keeps everything as exact fractions and never converts to decimals.
That is not fussiness. 1/42 is exactly 1/42; 0.0238 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 course's own numbers:
$$\frac{1}{6} + \left(-\frac{1}{2}\right) = \frac{1}{6} - \frac{3}{6} = -\frac{2}{6} = -\frac{1}{3}$$
Two things happened at the end. 2/6 was simplified to 1/3 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 course 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 losing one is an easy error to make in this course. If you take nothing else from this section, take this: when you reach an equation like
$$-\frac{1}{2} + 3 B_2 = 0$$
the term with B_2 in it must be the negative of everything else, because the
whole thing adds to zero. So 3 B_2 = 1/2, and B_2 = 1/6. If you get
-1/6, you have dropped the sign at exactly the step where the course 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 course 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 = 2. The sum runs j = 0, 1, 2, and the
coefficients come from row n+1 = 3, which is 1, 3, 3:
$$\binom{3}{0} B_0 + \binom{3}{1} B_1 + \binom{3}{2} B_2 = 0$$
$$1 \cdot B_0 + 3 B_1 + 3 B_2 = 0$$
Substituting the two values the recurrence starts from, B_0 = 1 and
B_1 = -1/2:
$$1 + 3\left(-\frac{1}{2}\right) + 3 B_2 = 0$$
$$1 - \frac{3}{2} + 3 B_2 = 0$$
The first two add to -1/2. Check that yourself: it is the fraction practice
from section 1. So -1/2 + 3 B_2 = 0, and by the sign rule above,
B_2 = 1/6.
Now check it at n = 3. The sum runs j = 0, 1, 2, 3, and the coefficients
come from row 4, which is 1, 4, 6, 4:
$$1 \cdot B_0 + 4 B_1 + 6 B_2 + 4 B_3 = 0$$
Substituting B_0 = 1, B_1 = -1/2, the B_2 = 1/6 you have just found, and
B_3 = 0:
$$1 + 4\left(-\frac{1}{2}\right) + 6\left(\frac{1}{6}\right) + 0 = 1 - 2 + 1 = 0$$
Which checks out. Each further value takes one more row of the triangle and the same two steps: substitute what you already have, then solve for the one value that is left. The Bernoulli unit asks you to take that method one step further, without this page open, and that 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.