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371. We will now show that this formula is true, whatever be the value of m. There will be two cases :

1st. When m is positive and fractional. Assume

ši

" =

. Let x = r; then zárt, and <=*.

r m =

Also let y' = 4; then yo = u, and y=u.

By proper substitutions we have

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Now suppose x = y, then z = u; and since r and s are positive whole numbers, we have from (1)

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-u

2-U

Hence, the formula is true when the exponent is positive and fractional.

2d. When m is negative, and either integral or fractional. Suppose the exponent of x and y to be -m; we shall have x-.-y-m = "y-m (2cm-Y");

hence,

m

(xm - y

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х="

(1) -Y Now suppose x = y; then whether m be integral or fractional, we shall have, from the principles already established,

x
my-m =
-x-sm, and

hence,

=m.c

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-Y

= (---m) X (m2*--) = -mx

Hence, the formula holds true universally.

BINOMIAL THEOREM.

372. The Binomial Theorem has for its object the development of a binomial with any exponent, into a series. This theorem is expressed by an equation, called the Binomial Formula.

373. It is required to expand (a+x)" into a series, n being any. real quantity, positive or negative, entire or fractional.

We observe that

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by a", we shall have the expansion of a+x)".

Put z =*; then 1+ (1+x)".

Let us now assume the equation,

(1+x)" = A+Bz+Cz+Dr+Ez +.... in which A, B, C, D, etc., are independent of %. We are to find the values of these coefficients which will render equation (1) true for all possible values of z.

Suppose z = 0; then from equation (1), we have A = 1.
Hence, the assumed development becomes
(1+z)" = 1+Bz+ Cz? + Dz*+Ez*+....

(2) for all values of %. Put z = u; then

(1+u)" = 1+Bu+Cu+Du'+ Eu+.... (3) Subtracting (3) from (2), and dividing the result by zu, we obtain (1+x)"-(1+u)"

DI

(4)

B+0(**)+D

Let P= 1+%, and Q=1+u; then P-Q = U.

Equation (4) now becomes P"-2"

u B+ P-Q

(5) Now suppose z=u; then P=Q. And by the Residual Formula (370), we shall have

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+ B

Substituting these values in equation (5), we have

n(1+z) = B+2 Cz+3D2° +4E2+.... ) Multiplying both members of equation (6) by (1+z), gives n(1+z)" = B+20 | 2+3D | Z+4E | z'+....

+2C +3D Multiplying both members of equation (2) by n, gives n(1+z)" = n+nBz+nCzo+n Dzi+.

(8) Now by equating the second members of (7) and (8) we shall have an identical equation, because it may

be satisfied for

any

value of z. Therefore the coefficients of the like powers of z in equations (7) and (8) are equal, each to each (368, III), and we shall have

B=n;

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= n.

Therefore, the values of the coefficients are

A=1.
В:

n(n-1)
C-

2

n(n-1)(n-2)
D

2 3
n(n-1)(n--2)(n-3)

2 · 3 · 4

Substituting these values in (1) we have

n(n-1) n(n-1)(n-2) (1+x)" =1+nz+ za +

2

2 · 3

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and by restoring the value of z, which is

a'
2, n(n-1). x, n(n-1)(n—2).2c*
=l+na+
2

2 3

(1+)"

at

a

a

ait.... B)

or, finally, multiplying both members of () by a,

n(n-1) n(n-1)(n—2) (n+x)" = an +nan-x+ an2x2 +

anamt.... (c) 2

2 . 3

Equation (C) is the binomial formula, as it is usually written. It will be observed, however, that in the three equations, (a), (6), (), the coefficients, or the factors depending on n, are the same; and in practice, either (a), (6), or may be employed, according to the form of the binomial to be expanded.

374. By inspecting the general formula (), we perceive that in the expansion of a binomial in the form of a+x, the law of the exponents is as follows:

1.- The exponents of the leading letter in the successive terms form a series, commencing in the first term with the exponent of the binomial, and diminishing by 1 to the right.

2.- The exponents of the second letter form a series, commencing in the second term with unity, and increasing by 1 to the right.

And the law of the coefficients is as follows:

3.The coefficient of the first term is unity, and that of the second term is the exponent of the required power.

4.-If the coefficient of any term be multiplied by the exponent of the leading letter in that term ; and divided by the exponent of the second letter plus 1, the result will be the coefficient of the following term.

375. If we take the least factor in each of che successive coef: ficients of the expansion, commencing at the second, we have a de creasing series

m, (n-1), (-2), (3), etc., in which the common difference is unity.

Suppose n to be a positive integer, then the least factor in the numerator in the (n +2)d term will be (nn), or 0, and this term will disappear. But if n is negative or fractional, then no one of the factors, (n-1), (n-2), (n-3), etc., can be zero, and the expansion may be continued indefinitely. Hence,

1.- When n is a positive integer, the expansion of the binomial will be a finite series, the number of terms being n+1.

2.- When n is negative or fractional, the expansion of the binomial will be an infinite series.

APPLICATION OF THE BINOMIAL FORMULA.

376. Let us resume the equation,

n(n-1)

n(n-1)(n—2) (a+X)" = an+nan-x+ Fan-x2 +

- an x + ... (C) 2

2 3 If n be entire and positive, this formula will be an expression of involution, denoting some power of the binomial.

If n be fractional and positive, the formula will be an expression of evolution, denoting some root of the binomial.

If n be negative, the formula will express the reciprocal of some power or root of the binomial.

377. Since the binomial coefficients depend entirely upon the exponent n, they may be formed independently. To do this, we

n-1 na -2 have simply to commence with unity and multiply by n,

2 3 etc., continually.

1. Expand (a—x) into a series.
Here n =

6; hence,
The first coefficient is

1 = 1
" second 6

1x6 = 6 6 third

6X = 15 16 fourth "

15 X 4 = 20 16 fifth

20 X 15 « sixth

15x} = 6 " seventh

6x= 1 Since the odd powers of —are negative, we have for the literal factors of the terms,

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