To verify this by the Binomial Theorem, we have

(1 − 2x + 3x2 - 4x3 +, &c.) 2n

and if the coefficient of 4 in the Binomial expansion be found, it will be seen to agree with the result just obtained.

Examples LXXIII.

1. Find the term involving x2y3 in (x + y + z)5; abcde in (a + b +c+d+e)1o; of x2y3z13 in (x2 + xy + yz)°; a2bc in (a - b + c) 4 ; a2b2c2 in (a2 + b2 + c2· ac bc ab)3; x10 in (1

2. Find the coefficient of x12 in (1 2x + 3x2 4x3+5x4)9; of xin (1 + x + x2 +, &c., ad inf.)"; of the term which is independent

3. Show that if the coefficient of a2bc be known in the expansion of (a+b+c)", those of ab2c and abc2 are also known.

4. Find the number of terms involving xy in the expansion of

(x + x13μ3 + x3y3 + x)15; and those involving æ2 in

12

(1

+

5

5. Find by the Multinomial and verify by the Binomial Theorem the following coefficients :

Of x2 in (1 + 2x + 3x2 +, &c., ad inf.) — 1;

Of x in (1 + 3x + 6x2 + 10x3 +, &c., ad inf.) — 12;

Of x2+3 in (1 + nx + n2x2 +, &c. ad, inf.)p+3;

6. Find the middle term of (1 - a + a2 - a3) 12, and the pth term in

7. Find to six terms the square root of 1+ 2x + 5x2 - 7x3; also to six terms the cube root of (x3 − x + x − 1 − x-3), in descending powers of x.

175. In Art. (168) it was proved that if the equality

We shall now extend this proof to the case of two variables x and y.

Let A + B + Cy + Dx2 + Exy + Fy2+ Gx3 + Hx3y +, &c. = M + Nx + Pý + Qx2 + Rxy + Sy2+ Tx3 + Ux2y +, &c.

for all values of x and y, and where the variations of x and
pendent of each other,

Then shall A =
Let the equality be written in the form

are inde

M; B = N; C

=

P, &c.

Since this is true for all values of x and y, suppose y to retain the value 0, while x may receive any other values whatever. The equality becomes

...

A+ BxDx2 + Gx3 + = M + Nx + Qx2 + Tx3 + ... and this is true for all values of x.

Hence A = M; B

=

N; D = Q; G

=

T, &c. Art. (168).

Again, we may suppose x to retain the value of 0, while y receives any other values whatever.

Again, in the third line, suppose y to retain the value 1, while x receives any other values whatever; we then have

Ex+Hx2+... = Rx + Ux2 +, &c.

for all values of x; whence

E = R; H = U; &c.

176. The application of this theory will be shown in the following examples.

1. Resolve the expression

Let

Or,

the quantities to be determined.

(x

1

=

1)(x-2)

A(x-2)+B(x+1)
(x-1) (x-2)

1 = A (x-2)+ B (x 1).

Since x may receive any value whatever, let x = 1.

In this case, the above equation becomes

Next, let x = 2; then the equation becomes

Proceeding, as in the former examples, we have

4 = A(x+2)(x + 1)3 + B (x + 3) (x + 1)3 + C (x + 3) (x + 2) +, &c.

+

+

(x+3) (x + 2) (x + 1) x+3 x+2 x+1'

The values of A, B, E may now be found as in Ex. 1.

Note. Let the following two points be noticed in this example.

1o. When the Denominator contains a power of any factor, as (x+1)3, a partial fraction must be assumed for that power, and for

each lower power down to the first,

example.

C

D

E

(x+1)3 ( +1) +1

in the

2o. When the Numerator corresponding to the highest power has been found, let that fraction be transposed to the first side of the equation, and be combined with the quantity already there; the expression on the first side will then reduce, as the Numerator and De

nominator have a common factor; (x + 1) in the example.

This

process must be continued until only the first power of the said factor remains on the second side.

... 3x2+26x + 83 = A (x2 + 7x + 20) + (Bx + C) (x − 1).

Let x =

have

1; then, substituting for x its value on both sides, we

Note. When the Denominator of the given expression contains a quantity which cannot be resolved into simple factors, the Numerator corresponding to that quantity must be assumed to be a series of powers of x, the highest index of which is less by 1 than the highest index in the quantity itself. Thus, in example (3), the Numerator corresponding to the Denominator x2+7x+20 was assumed to be Bx+C; for a Denominator x4 +1, the Numerator must be assumed to be Bx3+ Cx2 + Dx + E, and for x2m+1 the Numerator must begin with x2m −1 ̧

Again, in the equality (a) of example (3), it was assumed that if the equality hold for two values of x, the coefficients of like powers of x on each side must be the same. And it may be assumed generally, that if an equality between two expressions of the mth degree in x hold for m + 1 values of x, then the coefficients of like powers of 2 on each side must be the same.