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Assume

(a-x)=A+Bx+Cx2+Dx3+Ex1+&c.

Clearing the equation of surds by squaring, a—x—A2+2ABx+(B2+2AC)x2+(2ÀD+2BC)x2+&c.

By transposition, (A2-a)+(2AB+1)x+(2AC + B2)x2+ (2AD+2BC)x+&c.-0. Therefore A2-a=0,

... A—a. 2AB—1—0, .. 2AB——1, ... B

2AC+B=0, .. 2AC-B2, .. C=

1

2A

2a'

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Whence (a-x)=a2{1

2a 8a3 16a5

&c.} Ans.

2AD+2BC=0, .. AD-BC, .. D—

6. Required the development of (a) by this method. (a2x2)=A+Bx+Cx2+Dy3+&c. ; then, by squar

Assume

ing each side, and transposing, we have

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A2+2ABx+2AC +B2 +1.

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thod. Here, since the first term of the series must contain x,

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above method. Ans. a+3x+4x2+7x3+11x*+18x3+29x+&c.

3. It is required to convert

1-x 1-2x-3x2

into a series by the

same method. Ans. 1+x+5x2+13x3+41x2+121x+365x®+&c.

Assume

1-x
-2x-3x

·A+Bx+Cx2-Dx3—Ex1+Fx3+Gx®+,&c,

This, cleared of fractions, and by transposition, we have
A+ Blx+ Ca2+ D23+ Ex2+ Fx + G|x°+&c.
-1-2A 2B 2C 2D 2E 2F
2F&c.
+1-3A 3B 3C 3D 3El
Therefore A-10;. A 1.

B-2A+1=0 B-2A-1—2—1=1;
C-2B-3A 0; .. C-2B+3A=2.2+3=5

D-2C-3B=0; ... D=2C+3B=2.5

+3.1

=0.

13

E-2D-3C-0; ... E=2D+3C-2.13 +3.5=26+15=41 F-2E-3D=0; F=2E=3D=2.41 +3.13-82+39=121

G-2F-3E=0 ;.·.G=2F=3E=2.121+3.41—242+123-365

1-x 1-2x-3x2

=1+x+5x2+13x2+41x+121x+365x+&c. Ans.

4. It is required to convert (1-x) into a series by the same

method. Ans. 1

X 22

2 2.4

მე-3

3.52 3.5.7.25

2.4.6 2.4.6.8 2.4.6.8.10

-

&c. 5. It is required to find, according to the above method, the several roots or values of x, in the equation x-6x+13x2—12x—a, by means of a quadratic equation.

2=

Ans. x={{±√(@+)}• Put xy+z; then by substitution and ordering the equation, the terms I have y+y3(4z-6)+y2(622-18z+13)+y(4z—6)X (z2-3x+2)+z-62132-12za. (1.) Let 42-6-0; or z; then the odd powers of y disappear from the (1), and it becomes, by substitution for z, its value in the other terms, y—ly—3—1, or y-ya+f. Hence, by completing the square, I have y—by2+a+4.. y2± (a+4), or y √{}±√(a+4)} ; hence x=y+z=£±√{}±√(a+4).

6. Convert

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3-x-6x2

1

1+5x

2

and

or

or

or

1

or

(1+x)2 (1—x)2—2x3

3,

and

each of the foregoing

1—2x—x2+2x2 1-2ax+x2? 1-x-6x2 1+2x' 1+x+2x2 1+6x+x2

(1-x)'

expressions by the method of indeterminate coefficients.

Answers: 1+2x+8x+28x+100x+356x+&c., or 1+2x+3x2+5x3+8x*+13x2+&c., or 1+3x+5x2+7x3+&c. or 3+5x+7x2+13x3+23x1+&c., or 1+2ax+(4a2—1)x2+(8a3—4a) 23+&c., or 1+6x + 12x2+48x3+120x3+&c., or 1 2-4x+8x2-16x3+32x1&c., or 1+2x+3x3+4x3+5x+6x3+&c. or 1+3x+7x2+13x3+25x1+51x3+103x®+&c., or 1+8x + 27x2+64x3+125x*+216x5+343xo.

8 Assume

1-x

-3x-2=A+Bx+Cx2+Dx2+Ex2+Fx

1-3x-2x

This, cleared of fractions, and by transposition, we have A + Bx+ Cx2+ Dx3+ Ex2+ F\x5+ G|x®+ Hx2 1x-3B 3C23-3Dx3E 3Fx 2A│x2-2Bx3-2C\xa—2D\x3—2E\xo—2F\x2

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... B = 3A −1 = 2,. C-3B-2A-0; .. C3B+2A =8 D-3C-2B-0; .. D=3C+2B D=3C+2B = 28 E-3D-2C-0; ... E =3D+2C = 100 F-3E-2D-0;:.. F =3E+2D = 356 .. G=3F+2E =1268

G-3F-2E-0;

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=1+2x+8x2+28x3+100x*+356x+12682° Ans

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=A+Bx+Cx2+Dx3+&c. ;

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Then multiplying each side by a' + b'x, and transposing, we

have

Aa' +Aa'

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x

212 + Da'

)

Fcb' } 23 +&c.

Aa' a=0, therefore A=

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a

a'.

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Of the Binomial Theorem.

116. The Binomial Theorem is a general algebraical expression, or formula, by which any power, or root of a given quantity, consisting of two terms, is expanded into a series; the form of which, as it was first proposed by Newton, being as follows:

m"

m

mm―n m—2n.
n 2n

(P+PQ)P[1+a+ m m —n, Q2 + m m — n

n

n 2n

or

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(P+PQ)=P+AQ+. 2n

m- -n.

3n

m-3n

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BQ+ ∙CQ+ CDQ +

m-2n

3n

4n

&c. where P is the first term of the binomial, Q the second term

m

divided by the first, the index of the power or root, and A, B, C,

n

&c. the terms immediately preceding those in which they are first found, including their signs + or -.

Which theorem may be applied to any particular case, by substituting the numbers or letters in the given example, for P, Q, m and n, in either of the above formulæ, and then finding the result according to the rule.

When the index of the binomial is a whole number, the series will terminate, as observed under the article Involution before treated on; but when it is a negative or fractional number, as in the following examples, the series will proceed on ad infinitum, and will become more convergent the less the second term of the binomial is with respect to the first.*

*This celebrated theorem, which is of the most extensive use in Algebra, and various other branches of analysis, may be otherwise expressed as follows:

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(2a) = { 1.

ma-x

n`a+x'

x

-)2 +

m m+n_m+2n, x
2n 3n a+x

n 2n a+x n

:)+

mm+n

m m+n m+2n a—x3 +

n 2n a+x' n 2n 3n, a+x And if the reciprocals of the same expressions be required, they

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1. It is required to convert (a2+2) into an infinite series.

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(a2+x)2=a+

+

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2a 2.4a3 2.4.6a5 2.4.6.8a 2.4.6.8.10a9

Where the law of formation of the several terms of the series is

sufficiently evident.

2. It is required to convert

1

(a+b) or its equal (a+b)~, into an

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2n a+x n 2n 3n a+x) + &c. }

'a+x2

m

n`a+x'

It may here also be observed, that if m, taken singly, be made to represent any whole, or fractional number, whether positive or negative, the first of these expressions may be exhibited under the more simple form

(a++x)TM—aTM+maTM-1x+

+

m(m—1)
1.2

+

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m(m-1) (m2){M(28—1) } aTM—¿

........ 1 2.3 4.

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.n

where the last term is called the general term of the series, be cause if 1, 2, 3, 4, &c. be substituted successively for n, it will give all the rest.

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