Substituting these values in equation (5),

n (1+z)"−1 = B+2Cz+3Dz2+4Ez3+....

Multiplying both members of equation (6) by (1 + %),

n (1+z)" = B+2C\z+3D|z2+4E\z3+.

+ B +20 +3D

...

Multiplying both members of equation (2) by n,

(8).

n (1+z)" =n+nBz+nCz2+nDz3+.... . . . 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 (7) and (8) are equal, each to each (368, III), and we shall have

E=

2

1)

;

n (n − 1) (n − 2);

2

3

3

4

n (n − 1) (n − 2) (n − 3); etc.

and by restoring the value of z, which is a'

x n(n−1) x2 n(n−1)(n—2) x3

(1 + 2) " =

=1+n+

a

ast.....

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

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

374. By inspecting the general formula (c), 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 the successive coefficients of the expansion, commencing at the second, we have a decreasing series,

n, (n-1), (n-2), (n-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 (n — n), 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,

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 have simply to commence with unity and multiply by n, 1 n 2 etc., continually.

n

2

3

Since the odd powers of are negative, we have for the literal

factors of the terms,

ao, —α3x, -+ aax2, —a3x3, +a2x2, —αx3, +28.

Therefore the expansion will be

(a—x) = aε-6a5x+15a2x2-20a3x+15a2x-6x+x6.

2. Expand (a + x) into a series.

=

In this example n. Represent the coefficients by A, B C, D,....; then

at, atx, a ̄‡x2, atx, a txt, a $x,....

or by taking out the factor a, in the second member,

(a + x)3 =

a‡ (1 + þa ̄1x—

or by clearing of negative exponents,

....

2.4

We might have obtained this last result directly, by putting

the binomial in the form of a3(1 + 2) + .

to note the transformations made above.

It is well, however,

1

(a + x)2

= (a + x)~2 = a 2 (1 + 2) = 1/2(1 + 2)22.

=

a2

we obtain

5x4

+

+

α

a2

a3

a4

...).

4. Expand (a3-22)5 into a series.

If we take the descending powers of a3, commencing with the 5th, and the ascending powers of 2, commencing with the first, we have for the literal factors of the terms,

a15, a12x2, a3x1, aox, a3x3, x10.

Hence, with the coefficients the development becomes

=

(a3 — x2)5 — a15 — 5a12x2 + 10a3x — 10aox6 + 5a31⁄23 — x10.

Ans. 1 + 6c + 15c2 + 20c3 + 15c1 + 6c3 + co.

3. Find the seventh power of x + y.

Ans.

x2+xby+21x3y2+35x1y3+35x3y1+21x3y3 +7xy°+y7.

4. Find the eighth power of a2 — 1.

Ans. a16-8a14+28a12-56a10+70a3—56a6 +28a1—8a2+1. 5. Find the ninth of a power

Ans. a9— 9a°c + 36a2c2 — 84aoc3 +126α3c1 — 126a4c5 + 84a3c® . 36a2c+9ac8 - c9.

Ans. 15ax + 10a2x2 + 10a3x3 + 5a2x2 + α3x3.