These equations, for whatever section, are able to be chosen from the following Abacus [Table] which on account of its benefits both many and remarkable I am accustomed to call Pagcrhtos [Abacus Panchrestus: the 'good for anything' table].
The abacus is in fact perpendicular lines in distinct columns with the signs above of the letters A B C D and with the characters of the numerical figures or powers.
Among these numbers those [other numbers] being called Diagonals which are nearby [but] not in the same column.
All of these numbers are made by the addition of the Diagonals, of which the sum always being placed in the lower place near the same Column as the Diagonal, from the right margin more removed. Any number whatever is from its own Diagonal, by increasing towards the left, as of the vertical first to the margin second.
The numbers in Column A, are to their own diagonals in B ascending, as 2 to the marginal [number] of the second. [Thus, 5: 10 as 2: 4, or 10 = 5 4/2; 15:105 as 2: 14, or 105 = 15 14/2; etc]
Hence it follows the numbers being adjacent to the right margin and for the rest in succession the nearest can be found and being continued as far as to where it will have been seen; it is allowed that the whole abacus shall not be written down in the chapter. As beginning from the number 23 in the margin there will be the proportionals 184.108.40.206. That fourth [number] is of that diagonal as 23 [is to 2], which being adjacent to the diagonal gives 276, being written below in the same column. Then, 220.127.116.111 are proportionals. [In an inductive way, 253 = 23 22/2, 1771 = 253 21/3 = 23 22 21/3 2 1] To this fourth proportional being put near 253 shall become 2024 being written below in the same Column C. [In this case, 2024 = 276 22/3 = 24 23 22/3 2 1. Thus, Briggs has set out the two main generating properties for the binomial coefficients of Pascal's Triangle: nCr + nCr-1 = n+1Cr; and nC2 = n(n-1)/2, nC3 = n(n-1)(n-2)/3.2.1, etc].
 Of these numbers the usefulness is manifold: the first of which will be the finding of the usual numbers which we use in the generation or the analysis of the numbers of the figures [i.e. diagrams]. However the generation is with the given root by the addition of the gnomon to find the whole power; on the other hand, with the analysis with the power given by reckoning with the same gnomon we may wish to extract the same root.
These nearby numbers [in Table 8-2] have been placed ascending towards the left, of which the first and the last show the distance from unity of that power [by this Briggs means the magnitude of the power], which they serve. As 2 for the Square; 3, 3 the Cube; 4, 6, 4 the Biquadratic; 5, 10, 10, 5 for the complete or fifth power, etc.
The other use ought not to be valued less, has been for the finding of equations, with the help of which the Chords being acquired for all sorts of sections to the Periphery. And particularly for any section for which the parts are equal being given with an odd number of parts; with which the Chords themselves in a single operation being found.
Also from the same Table the Equations being obtained if the equal parts of the Periphery shall be with an even number of parts. These Equations truly do not present their own Subtended Chords with the single operation, but of Subtended Squares: Then from the Squares being given the Subtended Chords.
Powers for Equations having been placed one by one at the head of these Columns, each with the sign of Addition or Subtraction.
The numbers to be added for the powers being had by the addition of two nearby [numbers]in the same column.
And for these Equations the first number for the Chords, is that of the section number itself in Column A. The following in C. The third in E. And the rest by the same method, with other Columns being placed between.
All these numbers obliquely ascend towards the left. As in which place the number of the part is odd; for Trisection 3 - 1. For Quinquisection, 5 - 5 + 1. For septi-section 7 - 14 + 7 - 1. Which have been shown by me before. For the remainder of the sections with odd parts the method will be the same and being shown, with that usefulness before: and with being found from the Table. Thus, if the Arc of the Periphery being cut in 45 equal parts, the equation will be 45 - 3795 + 95634 - 1138500 + 7811375 - 34512075 + etc.
This whole Equation being taken from Adranus Romanus [ Adriaan van Roomen: see H. Goldstine, A History of Numerical Analysis..., Springer-Verlag, page 33, for a discussion of this famous equation proposed by Roomen, and solved by Vieta in a devastating manner] .
If the parts shall be 11, the equation will be 11 -55 + 77 - 44 + 11 - 1.
But for the Equations of Squares of Chords, the first number is the Square of the number of the section being sought in Column B. The second in Column D. The third in Column F. The fourth if H, etc.
As for the Bisection 4 - 1. For the Trisection 9 - 6 + 1. For the Quadrisection 16 - 20 + 8 - 1 . For Quinqui-section 25 - 50 + 35 - 10 + 1.
 It was also possible to be preparing this second Table, in which the single numbers are made as before by the addition of the Diagonals; and so as all kinds of Chords as Equations of squares, having less trouble than before. But any in particular of which the table describes without the continuation of the numbers from the top and head is much more difficult than with the previous table. There are still other uses of the first Table, which being unable to have [explained] here. For these reasons, I rate [these uses given ] before more than these coming after. The position of the numbers for any Equation you wish is the same as in the other Table, and which with two of the numbers being expressed from before, here being propounded singly.
Therefore with any Subtended Arc being given , if the Subtended Arc of multiples is being sought, at first being sought as many powers of the given Chord by continued multiplication of the same Chord by themselves and their own products. Then the following characteristic equations of any section, by addition and subtraction the power being found the Chords of the multiplied Arcs themselves, or the Subtended Squares.
For these Chords themselves where the number of the section is Odd; we had the examples being propounded above: but if the section shall be Even, the Squares should be found, as we shall soon see.
 And first in Bisection, where four Square Chords of which the Arc being given as you please, being equal to the Square of the subtended Arc doubled, and to the Biquadratic of the given Chords. Which will be able to be shown thus:
Let BC, CD, DE be equal inscribed [lines], and BE being continued, in F, thus as EF, ED being equal: and drawing BD, DF. The Triangles BCD, BDF have equal angles. For EFD, EDF are equal by the construction, and by Prop. 5, Book 1 . And DEB, double of the angle EFD, by Prop. 32, Book 1 and double of the angle EBD, or DBC, by Prop. 33, Book 6. The Triangles BCD, BDF are therefore equal angled,
And, by Prop. 4, book 6, similar. And the sides BC, BD, BF are continued proportionals. And the Square BD, being equal to the mean of the extremes to the rectangle BC, BF by Prop. 17, Book 6. [For BC/BD = BD/BF]
But BC is 1; and BE, 3 - 1, as is evident from the trisection [For BC = BG = p, and BCG being similar to ABC, GC = p2 = HD; hence, AH = HG = 1 - p2 ; from AG/AC = HG/DC, it follows that HG = (1 - p2)p = p - p3. Hence, BE = 3p - p3]; and EF, from the construction, certainly equals the line BC. The total therefore BF, will be 4 - 1, which being taken by BC, makes the rectangle 4 - 1.
[ As noted above], equal to the Square of the line BD of the mean of the three [proportionals]. Therefore the Square of the line BD is 4 - 1.
That which had to be shown.
If the Periphery being cut in four equal parts, as OV, VY, YX, XS; the Square of the line OS thus being computed.
The square of the line OY ( as we showed above), being equal to four of the Square of the line OV, less the Biquadratic of the same OV. For the same reason, the Square of the line OS, being equal to four of the square of OY, less the Biquadratic of the same line OY. The Square of the line OY is 4 - 1,
Which going into itself makes the Biquadratic of the line OY; 16 - 8 + 1.
[OY2 = 4p2 - p4; OS2 = 4OY2 - OY4 = 16p2 - 4p4 - 16p4 + 8p6 - p8 = 16p2 - 20p4 + 8p6 - p8]
If the Periphery being cut in eight equal parts, and the Square of the line subtending four parts shall be 16 - 20 + 8 - 1, this Square multiplied into itself gives the Biquadratic:
256 - 640 + 656 - 352 + 104 - 16 + 1 .
Four Squares are 64 - 80 + 32 - 4.
By taking the Biquadratic the Square of the line OS is left.
64 - 336 + 672 - 660 + 352 - 104 + 16 - 1 .
By the same method the Square of the line subtending 16 equal parts is computed.
If the number of equal parts shall be odd, the Chords (as we found above) multiplied by themselves will give the same Square of the Chord.
As the Chord of the triple Arc being equal to three of the roots less the cube
3 - 1
9 - 3
-3 + 1
9 - 6 + 1 The Square subtending three equal Arcs.
This Square being multiplied by itself gives the Biquadratic:
81 - 108 + 54 - 12 + 1.
Now four of the Square are 36 - 24 + 4.
Therefore by taking the Biquadratic there will remain
36 - 105 + 112 - 54 + 12 - 1.
So this will be the Square subtending six equal parts.
 By the same method the Squares being found of the lines subtending the arcs which being any multiple of the given arc. And if the Chord shall be given of any Arc you please for the parts [i.e. fractions] of which the Radius is 100000, the Square of the Arc for any multiples of the Chord will be able to be found from the same parts.
For the Powers of Chords should be prepared as many as will have been seen, and by the Multiplication, Addition, and Subtraction of these; according as for some equation will have been put in order, it will be easier to calculate the Chord itself or the Square of any multiple Arc: As being taken the Chord of 36 Degrees, which being equal to the larger segment of the Radius proportionally cut. Of which any Powers you wish being calculated by Subtraction alone, as you see here.
 If the Square of the nine times the Arc of the Chord is being sought or of 324 Degrees,
81 - 540 + 1386 - 1782 + 1287 - 546 + 135 - 18 + 1
For which duly by adding and subtracting there will remain the Square of the Chord 324:0'; or 36:0': 038196601125010515176.
If the Square of double the Chord is sought, indeed of 72:0':
If the Square of the quintuple of the Chord is sought, indeed of 180:0':
 The Chord of 12:0': being taken and of this several powers.
With these continued proportionals, if being given five together, and from the largest and five times the forth being taken five times the second, the remainder will be equal to the sixth. Thus this series may be continued as the use of which will have been seen. For by being given ABCDE, F being sought.
[ This amounts to saying that if p6 = 26 sin6 p/60 is the first term, then
p6 + 5p9 - 5p7 = p11, or 1 + 5p3 = 5p + p5: see Note 1 of Chapter 5, where A = 1 is the length of the quintuple of the chord].
If the Square of the Chord of the sextuple, namely of 72:0': degrees be required, the equation is 36 - 105 + 112 - 54 + 12 - 1 .
This equation accordingly being solved will give Square of the Chord of 72 Degrees
[Compare with the true value:
Which clearly agrees with the Square of the Radius and the sides of the inscribed Hexagon, as you see here.
 The Chords of 20. 100. 156. 84 follow together with several Powers of the same of which series, hardly being able to continue without difficulty and which will have been seen in any place: as [this]will be evident individually.
The Chord of 20:0' being taken with a few powers of the same as you wish.
If the Square of the Chord of six times the Arc, surely of 120:0' is sought: the equation will be 36 - 105 + 112 - 54 + 12 - 1 .
Let the given Chord be100:0':
For these continuing proportionals with being given three together it will be possible to find the fourth: For as before the first A taken from three of the second leaves the fourth D.
Let the given Chord be140:0': and Powers of the same.
If the first should be added to three of the Second, the total will be equal to the fourth
Let the given Chord be156:0': and Powers of the same.
With five closest neighbours given, if 5 of the Second should be taken away from five of the fourth and the first, the remainder will be the equal of the sixth.
Let the given Chord be 84: and Powers of the same.
Given 5 neighbouring [terms], if the First and five of the Second should be taken from five of the Fourth, the remainder will be the Sixth.
1 Table 8-2 contains the Binomial coefficients, yet it is not exactly Pascal's triangle, and has great versatility. It is hence a useful exercise to locate these coefficients within the table, a part of which is reproduced here:
The numbers in the second Table are formed from the vertical sums of two adjacent numbers in the same column above: We now have an extended note concerning the connection of these numbers with the coefficients in the section equations:
121 = 55 + 66 = 11C9 + 12C10 , etc.
Now, the odd sections have expansions such as :
7p - 14p3+7p5- p7,
which become in terms of binomial coefficients:
( 4C3 + 3C2 )p - (5C2 + 4C1) p3+ (6C1 + 5C0) p5- p7 ;
again, for the 11th power section:
11p - 55p3+77p5- 44p7 + 11p9- p11,
(6C5 + 5C4)p - (7C4 + 6C3)p3 + (8C3 + 7C2) p5- (9C2+ 8C1)p7 + (10C1+ 9C0)p9- p11.
Hence, we surmise that for the odd section of order M = 2N + 1, the governing equation will be:
(N+1CN + NCN-1)p - (N+2CN-1 + N+1CN-2)p3 + (N+3CN-2 + N+2CN-3) p5 - ... + (-1)r(N+r+1CN-r + N+rCN-r-1)p2r+1 + ... + (2NC1+ 2N-1C0)p2N-1 - p2N+1.
Thus, Adriaan van Roomen's famous equation of the 45th degree has N = 22, and the general term is of the form:
(-1)r(23 + rC22-r + 22+rC21-r)p2r+1:
For example, setting r = 3 gives
-(26C19 + 25C18)p7 = - (657800 + 480700) = - 1138500.
We should be able to derive these equations starting from De Moivre's Theorem:
According to which, for any given angle , usually taken with 0 < < 2p ,
ei = cos + i sin,
eiM = (cos + i sin)M = cos M + i sin M.
In the present circumstances, a chord of length A in a circle of radius R subtends an angle 2M, where A = 2R sin M, and where is odd as above, and equals 2N + 1. Hence we may write De Moivre's Theorem in the form:
2N+1C0 cos2N+1 + i 2N+1C1 cos2N sin - 2N+1C2 cos2N-1 sin2 - i 2N+1C3 cos2N-2 sin3 + ...
+ 2N+1C2r-2 cos2N-2r+3 sin2r-2 + i 2N+1C2r-1cos2N-2r+2 sin2r-1 -
2N+1C2r cos2N-2r+1 sin2r - i 2N+1C2r+1cos2N-2r sin2r+1 + ...
+ 2N+1C2N-2cos3 sin2N-2 + i 2N+1C2N-1cos2 sin2N-1 - 2N+1C2N cos1 sin2N
- i 2N+1C2N+1sin2N+1
= cos(2N+1) + i sin(2N+1).
sin(2N+1) = 2N+1C1cos2N sin - 2N+1C3 cos2N-2 sin3 +....
+ 2N+1C2r-1cos2N-2r+2 sin2r-1 - 2N+1C2r+1cos2N-2r sin2r+1 + ...
+ 2N+1C2N-1cos2 sin2N-1 - 2N+1C2N+1sin2N+1
= 2N+1C1(1 - sin2)N sin - 2N+1C3(1 - sin2)N-1 sin3 +....
+ 2N+1C2r-1(1 - sin2)N-r+1 sin2r-1 - 2N+1C2r+1(1 - sin2)N-r sin2r+1 +
+ 2N+1C2N-1(1 - sin2 ) sin2N-1 - 2N+1C2N+1 sin2N+1
= 2N+1C1 sin(1 - NC1sin2 + NC2sin4 - NC3sin6 + ... + (-1)N NCNsin2N)
- 2N+1C3 sin3 (1 - N-1C1sin2 + N-1C2sin4 - N-1C3sin6 + ... + (-1)N-1 N-1CN-1sin2N-2) + ...
2N+1C2r-1 sin2r-1(1 - N-r+1C1sin2 + N-r+1C2sin4 - N-r+1C3sin6 + ... + (-1)N-r+1 N-r+1CN-r+1sin2N-2r+2)
- 2N+1C2r+1 sin2r+1(1 - N-rC1sin2 + N-rC2sin4 - N-rC3sin6 + ... + (-1)N-r N-rCN-rsin2N-2r) + ...
+ 2N+1C2N-1 sin2N-1(1 - 1C1sin2) - 2N+1C2N+1 sin2N+1.
The terms are now gathered as a finite expansion of powers of sin, taking R = 1, where is the half-angle subtended by the chord: then
A = 2 sin(2N+1) =
sin . 2(2N+1C1) - sin3 . 2(2N+1C1 NC1 + 2N+1C3) + sin5 . 2(2N+1C1 NC2 + 2N+1C3 N-1C1 + 2N+1C5)
- sin7 . 2(2N+1C1 NC3 + 2N+1C3 N-1C2 + 2N+1C5 N-2C1 + 2N+1C7) - ...
+ (-1)r . sin2r+1 . 2(2N+1C1 NCr + 2N+1C3 N-1Cr-1 + 2N+1C5 N-2Cr-2 + ... + 2N+1C2r-1 N-r+1C1+ 2N+1C2r+1) + ...
+ sin2N-1 . 2(2N+1C1 NCN-1 + 2N+1C3 N-1CN-2 + 2N+1C5 N-2CN-3 + ... + 2N+1C2N-3 2C1+ 2N+1C2N-1) -
sin2N+1 . 2(2N+1C1 NCN + 2N+1C3 N-1CN-1 + 2N+1C5 N-2CN-2 + ... + 2N+1C2N-1 1C1+ 2N+1C2N+1)
If we identify these coefficients with those present in Briggs' second Abacus: for the linear and cubic terms, we have:
2(N+1CN + NCN-1)= 2(N+1C1 + NC1)= 2.2N+1C1 = 2.(2N + 1),
23(N+2CN-1 + N+1CN-2) = 23(N+2C3 + N+1C3) = 23(2N + 1)(N + 1)N/3!,
2(2N+1C1 NC1 + 2N+1C3) = 2(2N + 1)N + 2(2N + 1)2N(2N - 1)/3! = 23(2N + 1)(N + 1)N/3!.
For the (2r + 1)th term, we have
22r+1(N+r+1CN-r + N+rCN-r-1) = 2(2N+1C1 NCr + 2N+1C3 N-1Cr-1 + 2N+1C5 N-2Cr-2 + ... + 2N+1C2r-1 N-r+1C1+ 2N+1C2r+1);
For the last term, consider
2(2N+1C1 + 2N+1C3 + 2N+1C5 + ... + 2N+1C2N-1 + 2N+1C2N+1) .
22N+1 = (1 + 1)2N+1 =
2N+1C0 + 2N+1C1 + 2N+1C2 +... + 2N+1C2N-1 + 2N+1C2N + 2N+1C2N+1 ,
02N+1 = (1 - 1)2N+1 = 2N+1C0 - 2N+1 C1 + 2N+1C2 - ... -2N+1C2N-1 + 2N+1C2N - 2N+1C2N+1;
22N+1 = 2(2N+1C1 + 2N+1C3 + 2N+1C5 + ... + 2N+1C2N-1 + 2N+1C2N+1).
From this investigation, we assert:
22r(N+r+1CN-r + N+rCN-r-1) = 2N+1C1 NCr + 2N+1C3 N-1Cr-1 + 2N+1C5 N-2Cr-2 + ... + 2N+1C2r-1 N-r+1C1+ 2N+1C2r+1
is a valid Binomial Identity, for 0 r N.
A Proof of the above Briggs Identity using the Lagrange Inversion Theorem
David M. Jackson ( Dept. of Combinatorics and Optimization, University of Waterloo, Ontario.), has been kind enough to provide an independent proof of the above Briggs-related identity. His proof makes use of a form of the Lagrange Inversion Theorem, (see e.g. E.T. Whittaker and G. N. Watson, A Course on Modern Analysis, 4th edition, C.U.P., 1950, p. 130.), adapted for the ring of formal power series.
First some notation: The ring of all formal power series in an indeterminate over the rationals is denoted by Q[]. The elements of this ring are infinite series that do not necessarily converge. If f Q[], then [n] f denotes the coefficient of n in f. The statement of the theorem for Lagrange inversion of Q[] now follows:
Let () be a formal power series in such that () 0. Let F() be another formal power series in . Then the functional equation w = t(w) has a unique solution. Moreover, if n is a non-negative integer, then
F(w) = F(0) + (tn/n)[n-1]F'()n(.
Note that from the functional equation, w(0) = 0, and the condition (l) 0 insures that is invertible.
Consider the functional equation T = x eT for T(x). By the theorem there is a unique series T(x) Q[[x]] that satisfies this equation, and [xn]T =(1/n)[n-1](d/d)en = nn-1/n!, hence T(x) = (nn-1/n!)xn, where F(w) = w in this simple case.
Again, if we have T5 = x eT, then [xn] = (1/n)[n-1](d/d5)en = (5/n)nn-5/(n-5)!, and T5(x) = 5 (1/n)(nn-5/(n-5)!)xn.
Now to the Briggs-related identity:
Let r be a non-negative integer, then N-iCr-i .2N+1C2i-1 = 22r(N+r+1CN-r + N+rCN-r-1.
Let r,N denote the summation on the left hand side of the statement. Then, on two applications of the binomial theorem,
r,N = ([tr-1](1 + t)N-i)([x2i+1](1 + x)2N+1) = [tr](1 + t)N ti/(1+t)i([x2i](1 + x)2N+1)/x =
[tr](1 + t)N((1+ (t/(1+t)))2N+1)/(t/(1+t))
on setting x = (t/(1+t))as x and t are arbitrary .
Whence, noting that (1+t)N = ((1+t))2N, and re-arranging:
r,N = [t2r+1](t + (1+t2))2N+1
on setting t2 as the new variable.
Let q = t + (1+t2). Then q Q[[t]], and q2 = 1 + 2t2 +2t(1+t2) = 1 + 2tq. Let q2 = 1 + w, then w satisfies the functional equation w = 2tq = 2t(1 + w) in Q[[t]], and r,N = [t2r+1](q)2N+1= [t2r+1](1 + w)N+1/2 .
Hence, in the Lagrange Inversion Theorem, we set w = 2t(1+w) = 2t(w) where (w) = 2(1 + w), and the required expansion F(w) = (1 + w)N+1/2, to give
r,N = [t2r+1](q)2N+1 = [w2r+1](1 + w)N+1/2 = (1/(2r+1))[2r](N+1/2)(1 + )N-1/2 22r+1((1+))2r+1 =
22r(2N + 1)/(2r + 1) [2r](1 + )N+r = 22r(2N + 1)/(2r + 1) N+rC2r = 22r(N+rC2r+1 + N+r+1C2r+1)
which is equivalent to the statement of the theorem.
If you are interested in this sort of thing, and would like to become more conversant with these remarkable short-hand methods, then Combinatorial Enumeration, by I. P. Goulden and D. M. Jackson, Wiley (1983) is a text well worth examining. Lagrange's Theorem in the single variable form is on page 17, with further generalisations following and other examples.