Demonstration of the Binomial Theorem for any Exponent. 244. It has been shown (Art. 61), that any expression of the form

mym, is exactly divisible by x-y,

when m is a positive whole number. That is,

The number of terms in the second member is equal to m; and if we suppose x = y, each term will become equal to xm-1: hence,

We propose to prove that the quotient will have the same form when m is negative, and also, when m is a positive or negative fraction.

First, when m is a whole number, and negative. Let n be a positive whole number, and numerically equal to m. Then,

хст

ост

= mxm-1, m being a negative number.

Second, let m be a positive fraction, or m = 응.

If we suppose x = y, we have v = u; and since positive whole numbers, we have

= mxm-1, m being a positive fraction.

Third, let m be a negative fraction, or

But since p is a negative integer, and q a positive integer, we have from what has preceded, after making x =

y, which

245. We are now prepared to find a general formula for the development of the binomial (a+b)m, in which the exponent m is positive or negative, and either integral or fractional.

In order to simplify the process, let us place the binomial

under the form

(a + b)m = [a (1 + —-—-) ]m = am (1 + 2)" (Art. 220).

a

m

If we find the development of (1 + 2)",

and then multiply

it by am, the product will be the development of (a + b)m. In order further to simplify the expression, let us make

then, the binomial to be developed will be of the form

(1 + x)m.

As this development must be expressed in terms of x, and known quantities dependent for their values on 1 and m, we may assume A+ Bx + Cx2 + Dx3 + Ex1 + &c.

(1 + x)m

=

(1),

in which the co-efficients A, B, C, &c., are independent of x, and functions of 1 and m.

Now, since this equation is true for any value of x, if we make x = 0, we have

(1) = A = 1.

Substituting this value in equation (1), we have

(2).

(1 + x)m =1+ Bx + Сx2 + Dx3 + Ex1 + &c. Since the form of the above development will not be changed by placing y for x, we may write

(1 + y) = 1 + By + Cy2+ Dy3 + Ey1 + &c. . . . (3). Subtracting equation (2) from (3), and dividing both members by xy, we have

y=v-u.

Make 1 + x = v, and 1+y=u; whence, x

Substituting these values in the first member of equation (4),

while the quotients in the second member become, respectively,

Substituting these values in equation (5), and we have m (1 + x)m−1 = + B + 2Cx + 3Dx2 + 4Ex3 + &c. . . . . (6). Multiplying both members of this equation by 1+x, and arranging the second member with reference to x, and we have m (1 + x)m = B + 2C | x + 3D | x2 + 4E | x3 + &c. + B + 20 + 3D

If we now multiply equation (2) by m, we have

m (1 + x)m = m + mBx + mCx2 + mDx3 + mEx1 + &c.

If we place the second member of the last two equations equal to each other, we shall obtain an identical equation. Then, placing the co-efficients of the like powers of x equal to each other (Art. 240), we have

Substituting these values of A, B, C, D, &c., in equation (2),

a development which is of the same form as the one obtained in Art. 203, under the supposition of m being a positive and whole number.

1

we make m=- it becomes (+ a)n or

n

n

3

1

x + a =

+...)

The fifth term can be found by multiplying the fourth by

REMARK.-If in this formula, we make n = 2, n = 3, n = 4, &c., the development will become the approximate square root, cube root, fourth root, &c., of the binomial (x + a); and by assigning values at pleasure to x and a as well as to n, we can find any root whatever of any binomial. If n is negative, or fractional, there will be no limit to the number of terms to which the series may be carried. Such a series is called an infinite series.