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in which a and b may represent the numerical coefficients of x and y. Now denote the numerical coefficients of the expansion by C,, C2, C3, etc. We shall then have (ax+by)" = Cyan + C2an-ly+ Ozan-*y*+Cqanmoy' +

' ...., in which Ci a",

b C:

1 a'

n

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8.a aa aa au

CA

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5

1. Find the fourth power of 5a+3x. In this example we have

4, 5, b and the coefficients of the expansion will be

с
54

625
C
625 : 4.

1500
1500 · } .

1350
C
1350 · .

540
C 540 1:=

81

Hence, (5a+3x)* = 625a*+1500aRx+1350a’xo+540ax: +812*, Ans.

2c 4x 2. Find the fourth

power

of

3 5
2
4

6 4 3 6 We have n= 4, a=

b

and 3 5

5 2 5 Hence the coefficients of the expansion are

(3) = si
C2 = 16:4:8=13
C = 138..f=128
CA= 28 } = 513
Cg = 5:19=295

Hence, 2c 4x 16

128 128 512 256

c** cx' + 3 5 81 135 75

625 375

a

Ca

128 135

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EXAMPLES FOR PRACTICE.

1. Find the fourth power of 2x+5y.

Ans. 16x*+160x*y+600x*y*+1000.xy' +625y. 2. Find the fifth power of 2a43x.

Ans. 32a°—240a*x+720aox—1080aRx+810ax*—243x'. 3. Find the sixth power of 3+4x'.

Ans. 729+5832x*+19440x* + 34560x® +34560x® +184322" + 4096.c".

3a 4r 4. Find the fourth power of +

4 5 Ans. a+ a'r ffart 1gfar+ *ffr.

2t 3r 5. Find the sixth

power

of z + z Ans. t + tor+ 30*r*+20tor + 136?r* +24&tr + r.

1 6. Find the fifth

power

of
4 5

1
Ans.

+

+ 1024 256 160 200 500 3125

m

mo

m'

m'

m

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380. The approximate value of a surd root may be obtained with much facility by expanding the root into a series.

Let a" represent that perfect nth power, which is next less or next greater than the given number, and let b denote the difference between this power and the given number. Then

a"+b, ora"-6, will express the given number. But we have

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Va*+6 = apit and "ar_ = a

a"
Developing the radical parts into series, we have
Vaa+b=
16 1 1
-n

1 1-n 1-2n 3
+
+

om + ....) (1) na"

n
2n
a'r.

n 2n 3n

a(1+

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The second members of these equations contain no radicals; hence,

Any surd may be developed into a series of rational terms; whence by summing the series, we may obtain approximately the indicated root.

b It should be observed that the smaller the fraction is, the more

an rapidly will the series converge.

1. Find the cube root of 76, to six decimal places.

The smallest fraction will result by taking the cube which is next less than 76, or 64; thus,

8/76 = V/64+12=401+1 We may now develop the radical part by equation (1), in which

b 3 :3, a=

an 16: To form the binomial coefficients we have the factors,

1 1

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3

n =

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1-5n
6п

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9

1-3n 2

1-on 17
4n
3
7n

21 We represent the successive terms by A, B, C, etc.; and to secure accuracy in the final result to the 6th place of decimals, we should

3

16

1

3 16

.

1

carry the computation in each term to the 7th place. Thus we have A

+ 1.0000000 B = +

+ .0625000 с

B

.0039062
D

с + .0004069
D

.0000508
F
itE

.0000069
F

.0000010 37:38G + .0000001 Algebraic sum,

1.0589559

4

3

16

G =
H =

3 16

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76 =

n

1-n

25

Whence,

4.235824+, Ans. 2. Find the 5th root of 25, to 6 decimal places. The most convenient fraction will result by taking that 5th

power which is next greater then 25, or 32 ; thus,

V25 V327 2V1—31E Equation (2) will now apply; and the operation will be as follows:

1 1

5
2

1-4n 19
2n
5

5n
1-2n 3

1-5n 4
3n
5
бп

5
1-3n 7 1—6n

29
4n
10

35
A

+1.0000000 }

437500
C

32
B

38281
D

c

5024 16:33:D

769
19:35E

128
G =
F

22
Н
G

4 Algebraic sum,

.9518272 =

V1-3 2

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B =

7 32 7

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Find the values of the following indicated roots, to the 6th lecimal place : 1. V9.

Ans. 2.080084. 2. V31.

Ans. 3.141381. 3. V100.

Ans. 4.641589. 4. V110.

Ans. 4.791420. 5. V297.

Ans. 3.122851. 6. V60.

Ans. 1.978602. 7. V4.

Ans. 1.319508. 8. V3275.

Ans. 5.047104. 9. V125.

Ans. 1.993235.

EXPANSION OF FRACTIONS INTO SERIES.

381. An irreducible fraction may always be converted into a series, by dividing the numerator by the denominator.

1 1. Convert into a series.

1+a

1 Observe that

itaa+1 Hence, there may be two ways of dividing; 1st.

2d.

1 1 1 1+a)1 (1-a+a

a+1)1

+

1 ita

1+

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