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(2x2 + 3y13)* = (2x2)1 +4(2x2)3 × (3y2) + 6(2x2)2 × (3y2)2 + 4(2x2) × (3y2)3 + (3y2)* = 16x + 96x'y' + 216x'y' + 216x2y + 81y*, as required.

3. To find the seventh power of ab - 2cd.

By the Table, the coefficients in the seventh power are 1 + 7 + 21 + 35 + 35 + 21 + 7+1; and using ab and -2cd as simple letters in a binomial, we get (ab — 2cd) = (ab)2 + 7(ab)° × ( − 2cd) + 21(ab)3 × ( − 2cd)2 + 35(ab)* × (− 2cd)3 + 35(ab)3 × ( − 2cd)* + 21(ab)2 × ( − 2cd)3 +7(ab) × (-2cd)+(-2cd)'=a*b*— 14abcd + 84a3b3c2d2 — 280a*b*c3Ã3 + 560a3b3c*d* — 672a2b2c3d3 +448abc'd' — 128c'd', as required.

Ex. 16. To find the n power of the binomial a+b; supposing n to denote any positive whole number.

By the notation of powers, we shall have the n" power of a+b expressed by (a + b)" = (a + b). (a + b) . (a + b) . . . to n factors (A).

If we multiply the a of the first factor of the right member of (A) by the a of the second factor, and the product by the a of the third factor, and so on to all the factors, we shall get a α a... to n factors = an (1), by the notation of powers, and a will be the first term of the sought power, supposing it to be arranged according to the descending powers of a.

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If we change n in (1) into n − 1, we get an-1 for the term in the product of any (n-1) factors of (A) which is independent of b; and multiplying this term by the b of the remaining factor of (A), we get a"-1b, which will be repeated n times, when we take every one of the (n-1) factors of (A) (which are not all the same in any two cases), and proceed as before; that is, we get a"-1b + un−1b + an−1 + etc., to n terms = na"-1b (2) for the second term of the required power. Again, if we change n in (2) into n 1, we get a"-b + a”-2b + a2¬2b + etc., to (n − 1) terms = (n-1)an-b for the term of the product of any (n − 1) factors of (A) which involves the first power of b, and multiplying this term by the b of the remaining factor of (A), we get a”-22 + a”¬?f2+ an-2b+ etc., to (n - 1) terms = (n − 1)an-b2; which will be repeated n times, when we take every one of the (n − 1) factors of (A), that are not all the same in any two cases, and proceed as before; that is, we shall get a"-f3 + an-2f3 +

an-272 + etc., to n(n-1) terms = n(n − 1)an-263, which has twice as many terms as it ought to have in the sought power; for (by the process) one of the terms has arisen from multiplying an-2 by the b of the first factor, and the product by the b of the second factor; and another term has resulted from the multiplication of a"-2 by the b of the second factor, and the product by the b of the first factor; consequently, the product of an-2 by the b of the first and second factors has been taken twice, when it ought to have been taken but once; and in the same way the b in any two factors of (A) may be shown to give two terms instead of one, which we ought to have; consequently we shall have a"-272 + an―22 + n(n − 1) an−272 (3), for

an-etc., to

n(n - 1)

1 2

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terms =

1 2

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another term of the sought power.

If we put n-1 for n in (3), and multiply by b, we get an-973 + an-373 + etc., to (n-1) (n − 2) terms (n-1)

(n

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2)

2

1

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2

1

an-33 for the term of the product of any (n − 1)

factors of (A), which involves 62, and that is multiplied by the b of the remaining factor; and proceeding in this way with every (n − 1) factors of (A), (the factors not being all the same in any two instances), the above equation will be repeated n times, or we shall have a"-33 + an-373 + etc., to n(n − 1)(n − 2) terms = ̧n(n − 1) (n − 2) an-33, which has

1

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three times as many terms as it ought to have in the required power; for the b in any three factors of (A) ought evidently to give but one term, whereas, by the process we have used, each b in the three factors gives a term; hence we deduce n(n − 1) (n − 2)

an-373 + an−373 + etc., to

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an-373 (4) for another term of the sought power.

If for n we put n − 1 in (4), and proceed in like manner, we shall find "(n − 1) (n − 2) (n − 3)

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of the required power, and so on; and it is evident if m +1 represents the number of any term, that we shall have

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expression of the term, which is called the general term of the expansion of (a + b)".

Hence, collecting the terms, we get (a + b)" = a” + na”-1

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(n − 3) 4

a”-1b1 + etc. (B), for the required power, whose law

of continuation is manifest; and it is easy to see that the number of terms in the right member of (B) is n + 1, and that the last term is b".

Since a and b are clearly arbitrary, we may put -b for

b, and (B) becomes (a - b)" = a" — na”-1b +

n(n − 1) (n − 2)

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n(n − 1)

1 2

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·an-373 + etc. (B′), observing the rule of

If in (B) we change b into b, we shall have (a ±b)" =

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(C), which comprehends (B) and (B') in a single formula, observing that for (which is called the ambiguous sign) we must use when b is to be taken as a positive quantity, and that must be used for

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when b is to be negative, or

It is clear that (C) is an identical equation, since its right member is the expansion of its left or first member; in other terms, its right member is the same as (a ±b)" under a different form.

Hence, although the equation has been rigorously established, only on the supposition that n is a positive whole number, yet its right member ought evidently to be considered as the development of its first member, or of (a ±b)", whatever may be the value of n; that is, whether n is integral or fractional, rational or irrational, positive or negative,

etc.

Regarding b as an increment of a, and b as a decrement of a, we may clearly derive (C) from the identical equation a" = a", in the following manner.

By adding to a, the first member of the equation. becomes (ab)"; and to get the termna"-1b from a", we subtract unity from its exponent, which reduces it to an-1; then we multiply together the exponent n of the power a", a"-1, and b, which gives na"-1b, as required.

To get the term

n(n − 1)
1 2

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a"-2b2, we resume the term ±

na"-1b, then, subtracting 1 from n - 1, the exponent of a in the term, it is reduced to na"-2b; and multiplying together the index n 1 of a, in the term, na"-2b, and ±b, and dividing by the index of b in the product, we get n(n − 1) 1 2

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an-22, as required. In a similar way we derive

the term ±

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n(n − 1) (n − 2) an±373 from the preceding term

1 2

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3

n(n − 1) an-22; by diminishing the exponent of a by 1, and

1.2

then multiplying together the exponent n 2 of a,

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an-86 and b, and dividing by the exponent of b in the

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n(n − 1) (n 2)

an-4b4

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and so on, to any extent. Hence it is easily seen that the identical equation a"a", by adding

b to a, will be changed (by the process we have used) into (C).

It may be observed that (C), when n is taken in its most. general sense, is called the Binomial Formula or Theorem of Sir Isaac Newton, because it is supposed to have been discovered by him. And it may be added that the method which we have given of deriving any term of the right member of (C) from the next preceding term is substantially the same as the rule given by Newton for the development of a binomial.

We shall now give some simple examples to show the use of (C).

1. To find the third power of 7ab - 3c.

In this question we must put n = 3, and use forin (C); which, being done, (C) becomes (a - b) = a3 3a3-16 3(3—1) (3 — 2)

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a3-373; or, since 3-1=2,

+

3(3-1) 1 2

a3-272

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a3 3 = ao = 1, we get (a - b) = a3 — 3a2b+3cb2 — b3.

To find the third power of 7ab - 3c, we must use 7ab and 3e for a and b in the preceding equation, and we get (7ab 3c)3 = (Tab)3 — 3(7ab)2 × 3c + 3(7ab) × (3c)3 — (3c)3 = 343a3b3 ·441a2bc+189abc2 — 27c3, as required.

2. To find the sixth power of 1 + 3x.

In this question we must put n=6, and use + for in (C), and it reduces to (a + b) = ao + 6ab + 15a1b2 + 20a3b3 +15a2b1 + Cab3 +b6; hence, using 1 for a, and 3x for b, in this equation, we get (1 + 3x) = 1 + 6 × 3x + 15 × (3x)2 + 20 × (3x)3 + 15 × (3x)1 + 6 × (3x)3 + (3x)° = 1 + 18x + 135a2 + 540x3 + 1215æ1 + 1458æ3 + 729x, as required. If we put x = 1, then 1+3x= 4, and we have 46 1 +18+ 135+ 540 + 1215 + 1458 + 729 = 4096, as it ought to do. 3. To find the fourth power of

x

y2 3 4

In this question we must put n = 4, and use for ± in (C), and it changes to (a - b) = a1 — 4a3b + 6a2b2 — 4ab3 + ¿1; and putting & for a,

3

3

y2

4

( − 4)* = ( )* − 4 ()'(4)

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81

=

1, y

+

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+6

ys +

27 24 48 256

=

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1, our result reduces to

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as it ought to do.

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We shall conclude Multiplication by giving a few examples for practice.

EXAMPLES.

1. Multiply x2 + xy + y2 by x2 — xy + y2.

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