=

Any particular case is readily reduced to this formula; for making n = 2, and observing, that when there is only one letter, ૨ 1, we have 1 × 2 = 2 for the number of arrangements of a product of two letters. Again, taking Q1 X 2 and n = 3, we have IX 2 X 36 for the number of arrangements of a product of three letters; further, making Q = 1 × 2 × 3 and n = 4, there result 1 × 2 × 3 × 4, or 24 possible arrangements in a product of four letters, and so on..

140. What we have now said being well understood, it will be perceived, that by dividing the whole number of arrangements obtained from m letters, taken n at a time, by the number of arrangements of which the same product is susceptible, we have for a quotient the number of the different products, which are formed by taking in all possible ways n factors among these m letters.

This number will, therefore, be expressed by

P (m

− n + 1)

Q n

;

*

which being considered in connexion with what was laid down in

am-n in the developement of (x + a)m.

It is evident, that the term which precedes this will be express

P

ed by an―1 xm―n+1; for in going back towards the first term, the Q

exponent of x is increased by unity, and that of a diminished by unity; moreover, P and Q are the quantities, which belong to the number n

1.

* It may be observed, that if make successively

numbers, which express respectively, how many combinations may be made of any number m of things, taken two and two, three and three, four and four, &c.

These results show how each term in the developement of (x + a)TM, is formed from the preceding.

Setting out from the first term, which is am, we arrive at the second, by making n = 1; we have M = 1, since x has only

unity for its coefficient; the result then is

m

a xm-1, or

— a am-1. In order to pass to the third term, we make M =

and n 2, and we obtain

found by supposing M =

m

I'

a2 xm-2. The fourth is

a3 xm-3, and so on; whence we have the for

To pass from one term to the following, we multiply the numerical coefficient by the exponent of x in the first, divide by the number which marks the place of this term, increase by unity the exponent of a, and diminish by unity the exponent of x.

Although we cannot determine the number of terms of this formula without assigning a particular value to m; yet, if we observe the dependence of the terms upon each other, we can have no doubt respecting the law of their formation, to whatever extent the series may be carried. It will be seen, that

expresses the term, which has n terins before it.

This last formula is called the general term of the series

m

m (m

1)

a xm-1 +

a2 xm 2 + &c.

1.2

n = 3, &c.

because if we make successively

n = 1, n = 2,

it gives all the terms of this series.

142. Now, if (x + a)5 be developed, according to the rule

given in the preceding article; the first term being

Here the process terminates, because in passing to the following term it would be necessary to multiply by the exponent of x in the sixth, which is zero.

This may be shown by the formula; for the seventh term, having for a numerical coefficient

same factor entering into each of the subsequent terms, reduces it

to nothing.

Uniting the terms obtained above, we have

(x + a)5 = x5 + 5 a x2 + 10 a2 x3 + 10 a3 x2 + 5 aa x + a3.

143. Any power whatever of any binomial may be developed by the formula given in art. 141. If it were required, for example, to form the sixth power of 2 3-5 a3, we have only to substitute in the formula the powers of 2 a3 and 5 a3 respectively for those of x and a; since, if we make

we have

(2 x3- 5 a3)6 = (x2 + a')ε =
x3

26 +6 α' x 5 + 15 a 2x4 +20 a 3x3

+ 15 a1x/2 + 6 a' 5 x + a' 6 (141),

and it is only necessary to substitute for and a' the quantities, which these letters designate. We have then

(223)6 + 6 (→ 5 a3) (2 x3)5 + 15 — 5 a3)2 (2 x3)4

1

+ 20 (− 5 a3)3 (2 x3)3 + 15 — 5 a3)4 (2 x3)2

The terms produced by this developement are alternately positive and negative; and it is manifest, that they will always be so, when the second term of the proposed binomial has the sign

144. The formula given in art. 141, may be so expressed as to facilitate the application of it in cases analogous to the preceding.

by insulating the common factor am. In applying this formula, the several steps are, to form the series of numbers,

to multiply the first by the fraction, then this product by the second

and also by the fraction, then again this last result by the third

x'

and by the fraction, and so on; to unite all these terms, and add unity to the sum; and lastly, to multiply the whole by the factor x. In the example (2 x3- 5 a3), we must write (23)6 in the place of x, and

5 a3
2 x3

a

in that of I shall leave the application

of the formula as an exercise for the learner.†

+ The formula for the developement of (x + a)m answers for all values of the exponent m, and is equally applicable to cases in

145. We may easily reduce the developement of the power of any polynomial whatever, to that of the powers of a binomial, as may be shown with respect to the trinomial a+b+c, the third power for instance being required.

First, we make b + c = m, we then obtain

(a+b+c)3 = (a + m)3 = a3 + 3 a2 m + 3 a m2 + m3 ; substituting for m the binomial b + c, which it represents, we have (a+b+c)3 = a3 + 3 a2 (b + c) + 3 a (b + c)2 + (b + c)3. It only remains for us to develope the powers of the binomial b+c, and to perform the multiplications, which are indicated; we have then

a3 + 3 a2 b + 3 a b2 + b3

+3a2c+6 a b c + 3b2c
+3 ac2+3bc2

+ c3.

Of the Extraction of the Roots of Compound Quantities.

146. Having explained the formation of the powers of compound quantities, I now pass to the extraction of their roots, beginning with the cube root of numbers.

In order to extract the cube root of numbers, we must first become acquainted with the cubes of numbers, consisting of only one figure; these are given in the second line of the following table;

J

1

2 3
8 27

4 64

5 6 7 8 9 125 216 343 512 729

and the cube of 10 being 1000, no number consisting of three figures can contain the cube of a number consisting of more than one.

The cube of a number consisting of two figures is formed in a manner analogous to that, by which we arrive at the square; for if we resolve this number into tens and units, designating the first by a, and the second by b, we have

(a + b)3 = a3 + 3 a2 b + 3 a b2 + 63.

which the exponent is fractional or negative. This property, which is very important, is demonstrated in a note to the last part of the Cambridge course of Mathematics on the Differential and Integral Calculus.