398. To resolve a fraction into partial fractions is to express it as the sum of a number of fractions of which the respective denominators are the factors of the denominator of the given fraction. This is the reverse of the process of adding fractions that have different denominators. Resolution into partial fractions may be easily accomplished by the use of indeterminate coefficients and the theorem of § 396.

In decomposing a given fraction into its simplest partial fractions, it is important to determine what form the assumed fractions must have. Since the given fraction is the sum of the required partial fractions, each assumed denominator must be a factor of the given denominator; moreover, all the factors of the given denominator must be taken as denominators of the assumed fractions.

Since the required partial fractions are to be in their simplest form, incapable of further decomposition, the numerator of each required fraction must be assumed with reference to this condition. Thus, if the denominator is

x" or (x + a)", the assumed fraction must be of the form

it could be decomposed into two fractions, and the partial fractions would not be in the simplest form possible.

When all the monomial factors, and all the binomial factors, of the form x±a, have been removed from the denominator of the given expression, there may remain quadratic factors that cannot be further resolved; and the numerators corresponding to these quadratic factors may each contain the first power of x, so that the assumed

fractions must have either the form

Ax + B x2 + ax + b

or the

Since x + 1 = (x + 1) (x2 - x + 1), the denominators will be x + 1 and x2 x + 1.

3

1

x-2

4 x3 - x2-3x

x2 (x + 1)2

The denominators may be x, x2, x + 1, (x + 1)2.

... 4 x3 - x2 – 3x - 2 = Ax (x + 1 3x - 2 = Ax (x + 1)2 + B (x + 1)2 + Cx2 (x + 1)+Dx2

= (A+C) x3 + (2 A + B + C + D) x2+(A+2B)x+B;

A + C = 4,

2A + B + C + D = − 1,

A+2B =- 3,

CHAPTER XXV.

BINOMIAL THEOREM.

399. Binomial Theorem, Positive Integral Exponent. By successive multiplication we obtain the following identities: (a + b)2 = a2 + 2ab + b2;

(a + b)3 = a3 + 3a2b + 3ab2 + b2;

(a + b)2 = a2 + 4 ab + 6 a2b2 + 4 ab3 + b2.

The expressions on the right may be written in a form better adapted to show the law of their formation :

NOTE. The dot between the Arabic figures means the same as the

sign X.

400. Let n represent the exponent of (a + b) in any one of these identities; then, in the expressions on the right, we observe that the following laws hold true:

1. The number of terms is n + 1.

2. The first term is a", and the exponent of a is one less in each succeeding term.

3. The first power of b occurs in the second term, the second power in the third term, and the exponent of bis one greater in each succeeding term.

4. The sum of the exponents of a and b in any term is n. 5. The coefficients of the terms taken in order are 1; n;

n (n - 1). n (n - 1) (п – 2)

1.2

;

1.2.3

; and so on.

401. Consider the coefficient of any term; the number of factors in the numerator is the same as the number of factors in the denominator, and the number of factors in each is the same as the exponent of b in that term; this exponent is one less than the number of the term.

402. Proof of the Theorem. That the laws of § 400 hold true when the exponent is any positive integer is shown as follows:

We know that the laws hold for the fourth power; suppose, for the moment, that they hold for the kth power. We shall then have

(a + b)* = ax + kak-1b +

k (k-1)
1.2

ak-272

k (k-1) (k-2) + 1.2.3

ak-363+

ak-162

- .....

Multiply both members of (1) by a + b; the result is (k+1) k (a + b)x+1 = ax +1 + (k+1) ab + 1.2 (k+1) k (k-1) 1-2-3 In (1) put k + 1 fork; this gives

+

(1)

ak-263 + ..... (2)

(a + b)k+1 = a2+1+(k+1) a*b + (k+1) (k + 1 − 1) −172

+

1.2 (k + 1)(k + 1 − 1) (k +1 -2) 1.2.3

= a*+1 + (k + 1) akb +

ak-278 + .....

(k+1) kak-12

1.2

Equation (3) is seen to be the same as equation (2).

(3)