ao, —aax, +a*x3‚ —a3x3, +a3x1‚ —ax°, +x®.

Therefore the expansion will be

(a-x)—a—6ax+15a*x2-20a x+15a3x-6ax®+x®.

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

In this example n=1. Represent the coefficients by A, B, C, D....; then

a1, a ̄x, a ̄2x2, a ̄2x2, a ̄‡x2. a ̄2x2,........

(a+x)‡ = a‡ (1+ za

or by taking out the factor a3, in the second member,

x

3

2.4.6

2.4.6.8

2a 2.4a2

+

We might have obtained this last result directly, by putting the

binomial in the form of

að (1+ 2 ) §.

It is well, however, to note

the transformations made above.

a

(a+x)→ = a~(1 + ~ ) ̄ = '12; (1+2) ̃

α

4. Expand (a3-x3)3 into a series.

If we take the descending powers of a3, commencing with the 5th, and the ascending powers of x2, commencing with the first, we have for the literal factors of the terms,

a1,

10

a", a12x2, a'x*, a°x, a'x, x1o.

Hence, with the coefficients the development becomes

(a3—x3) — a1-5a12x2+10x-10a x +5a*x®—x1o.

3. Find the seventh power of x+y.

Ans. x+7xy+21x'y'+35x'y'+35x*y*+21x'y'+7xy®+y'.

4. Find the eighth power of a2-1.

Ans. a16-8a1+28a1-56a1+70a3—56a°+28a*—8a2+1. 5. Find the ninth power of a-c.

Ans. a'-9a°c + 36a c3-84a c' + 126a*c*—126a*c* + 84a3c36a'c'+9ac-c'.

6. Expand (1+ax)'.

Ans. 1+5ax+10a3x+10a3x+5a*x*+a*x°.

7. Expand (a-x3).

Ans a1-6a1x+15a*x*-20a°x+15a*x*-6a2x1o+x1.

8. Expand (x3-z′)°.

Ans. x1o—5x3z*+10x®z*—10x*z12+5x3z1o—z3°.

9. Expand (ax+dy)".

Ans. a1x+6ax*dy'+15a'x'd'y'+20a°x'd'y°+15a*x*d*y*+ 6a2xd'y1+d'y1.

10. Expand (a-x) into a series.

Ans. a(1+2x+3x2+4x3 +5x+6x® + . . . . ).

16. Expand (a+b) into a series.

19. Expand (1—a)-3 into a series.

Ans. 1+3a+6a2 + 10a3 +15a* +21a3 +28a +36a2 +

20. Expand (a2—x2) into a series.

we may suppose x and y to represent any quantities whatever; and thus we may obtain the development of the powers of binomials with numerical coefficients, or of polynomials.

1. Involve 3a+2c to the fifth power.

The binomial coefficients for the fifth power are

1, 5, 10, 10, 5, 1.

And by connecting these with the powers of the given terms, according to the law of the formula, we have

(3a+2c)* = (3a)*+5(3a)*(2c)+10(3a)'( 2c)'+10(3a)1(2c)'+ 5(3a)(2c)*+(2c)*;

or, by performing the operations indicated,

(3a+2c)* = 243a +810a*c+1080a3c'+720a3c'+240ac*+32c

2. Involve a+b+2c to the fourth power.

We may consider the polynomial in two parts, a+b, represented by x, and +2c represented by y. Then we have

(a+b+2c3)* = (a+b)*+4(a+b)3 (2c3)+6(a+b)*(2c3)*+ 4(a+b)(2c3)3+(2c3)*.

Performing the operations indicated,

(a+b+2c3)*

=a*+4a3b+6a2b3+4ab3+b‘+8a°c3+24a2bc3+

24ab'c'+8b'c'+24a3c*+48abc*+24b3c*+32ac°+32bc®+16c®.

EXAMPLES FOR PRACTICE.

1. Find the third

power of a- --26.

Ans. a-6a2b+12ab'—8b".

2. Find the fourth power of 2a+3x.

Ans. 16a +96a3x+216a2x2+216ax*+81x*.

3. Find the fourth power of 1-ļa.

Ans. 1—2a+a*—1a'+1'¿a*.

4. Find the fourth power of a'-ax+x2.

Ans. a3 —4a'x+10a°x2 — 16a3x3 +19a*x* — 16a3x3 +-10a3x® —

4ax+x3.

5. Expand (4a2—3x) into a series.

379. When a binomial having numerical coefficients is to be raised to any power, the coefficients of the expansion may be obtained with great facility by means of a simple modification of the binomial formula. We have (z+u)” =