P

Q

which precedes it, multiplied by m-n+1, the exponent of x in that term, and divided by n, the number of terms preceding the required term. This law serves to develop a particular power, without our being obliged to have recourse to the general formula.

For example, let it be required to develop (x+a). From this law we have,

(x+a)=x+6ax3+15a2x2+20a3x3+15a2x2+6a3x+ao.

After having formed the first two terms from the terms of the general formula "+max-1+. . ., multiply 6, the co-efficient of the second term, by 5, the exponent of x in this term, then divide the product by 2, which gives 15 for the co-efficient of the third term. To obtain that of the fourth, multiply 15 by 4, the exponent of x in the third term, and divide the product by 3, the number of terms which precede the fourth, this gives 20; and the co-efficients of the other terms are found in the same way.

In like manner we find

(x+a)100+10ax +45a2x2+120a3x2+210ax®,

+252a5x5+210aox2+120ax3+45a3x2+10a3x+a1o.

204. It frequently occurs that the terms of the binomial are affected with co-efficients and exponents, as in the following example. Let it be required to raise the binomial 3a2c-2bd to the 4th power.

Placing 3a2c=x and -2bd-y, we have

(x+y)*=x*+4x3y+6x3y2+4xy3+ya.

Substituting for x and y their values, we have (3a3c-2bd)*=(3a2c)*+4(3a2c)3(—2bd)+6(3a2c)2(−2bd)2+ 4(3a2c) (−2bd)3+(−2bd)1,

or, by performing the operations indicated

(3a3c-2bd)*=81a3c1-216a°c3bd+216a2c2b3d2—96a2cb3d3

+16b+d+.

The terms of the development are alternately plus and minus, as hey should be, since the second term is

205. The powers of any polynomial may easily be found by the binomial theorem.

For example, raise a+b+c to the third power.

Or, by substituting for the value of d,

(a+b+c)3=a3+3a2b+3ab2+b3

3a2c+3b2c+6abc

+3ac2+3bc2
+ c3.

This expression is composed of the cubes of the three terms, plus three times the square of each term by the first powers of the two others, plus six times the product of all three terms. It is easily proved that this law is true for any polynomial.

To apply the preceding formula to the development of the cube of a trinomial, in which the terms are affected with co-efficients and exponents, designate each term by a single letter, then replace the let. ters introduced, by their values, and perform the operations indicated. From this rule, we will find that

(2a2-4ab+3b2)3=8a6-48a5b+132a+b2-208a3b3

+198a2b-108ab5+27b.

The fourth, fifth, &c. powers of any polynomial can be developed in a similar manner.

Consequences of the Binomial Formula.

may

206. First. The expression (x+a)" being such, that be substituted for a, and a for x, without altering its value, it follows that the same thing can be done in the development of it; therefore, if this development contains a term of the form Karam—", it must have another equal to Kx"a"-" or Ka”—"x". These two terms of the development are evidently at equal distances from the two extremes; for the number of terms which precede any term, being indicated by the exponent of a in that term, it follows that

the term Kam-n has n terms before it; and that the term Kam—"x" has m-n terms before it, and consequently n terms after it, since the whole number of terms is denoted by m+1.

Therefore, in the development of any power of a binomial, the coefficients at equal distances from the two extremes are equal to each other.

REMARK. In the terms Ka"-", Ka"-"x", the first co-efficient expresses the number of different combinations that can be formed with m letters taken n and n; and the second, the number which can be formed when taken m―n and m-n; we may therefore conclude that, the number of different combinations of m letters taken n and n, is equal to the number of combinations of m letters taken m—n and

m-n.

For example, twelve letters combined 5 and 5, give the same number of combinations as these twelve letters taken 12-5 and 12—5, or 7 and 7. Five letters combined 2 and 2, give the same number of combinations as five letters combined 5-2 and 5-2, or 3 and 3.

207. Second. If in the general formula,

m-1

(x+a)=x+maxTM-1 ·+m- ·a2x2+, &c.

2

That is, the sum of the co-efficients of the different terms of the formula for the binomial, is equal to the mth power of 2.

Thus, in the particular case

(x+a)5=x2+5ax2+10a2x2+10a3x2+5aax+a3,

the sum of the co-efficients 1+5+10+10+5+1 is equal to 25 or 32. In the 10th power developed, the sum of the co-efficients is equal to 210 or 1024.

208. Third. In a series of numbers decreasing by unity, of which

the first term is m and the last m-p, m and p being entire numbers, the continued product of all these numbers is divisible by the con. tinued product of all the natural numbers from 1 to p+1 inclu. sively.

ber. For, from what has been said in (Art. 201), this expression represents the number of different combinations that can be formed of m letters taken p+1 and p+1. Now this number of combina, tions is, from its nature, an entire number; therefore the above expression is necessarily a whole number.

Of the Extraction of the Roots of particular numbers.

209. The third power or cube of a number, is the product arising from multiplying this number by itself twice; and the third or cube root, is a number which, being raised to the third power, will produce the proposed number.

The ten first numbers being

their cubes are

10.

1, 2, 3, 4, 5, 6, 6, 7, 8, 9, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. Reciprocally, the numbers of the first line are the cube roots of the numbers of the second.

By inspecting these lines, we perceive that there are but nine perfect cubes among numbers expressed by one, two, or three figures; each of the other numbers has for its cube root a whole number, plus a fraction which cannot be expressed exactly by means of unity, as may be shown, by a course of reasoning entirely similar to that pursued in the latter part of (Art. 118).

210. The difference between the cubes of two consecutive num. bers increases, when the numbers are increased.

Let a and a+1, be two consecutive whole numbers; we have

whence

(a+1)3=a3+3a2+3a+1;

(a+1)3—a3=3a2+3a+1.

That is, the difference between the cubes of two consecutive whole numbers, is equal to three times the square of the least number, plus three times this number, plus 1.

Thus, the difference between the cube of 90 and the cube of 89, is equal to 3(89)2+3×89+1=24031.

211. In order to extract the cube root of an entire number, we will observe, that when the figures expressing the number do not exceed three, its root is obtained by merely inspecting the cubes of the first nine numbers. Thus, the cube root of 125 is 5; the cube root of 72 is 4 plus a fraction, or is within one of 4; the cube root of 841 is within one of 9, since 841 falls between 729, or the cube of 9, and 1000, or the cube of 10.

When the number is expressed by more than three figures, the process will be as follows. Let the proposed number be 103823. 103.823 47

This number being comprised between 1,000, which is the cube of 10, and 1,000,000, which is the cube of 100, its root will be expressed by two figures, or by tens and units. Denoting the tens by a, and the units by b, we have (Art. 198),

(a+b)3=a3+3a2b+3ab2+b3.

Whence it follows, that the cube of a number composed of tens and units, is equal to the cube of the tens, plus three times the product