x in the same term, and dividing the product by 2. The coefficient of the fourth term is found by multiplying that of the third term by the exponent of x in the same term, and dividing the product by 3. For example, the second term of the development of (x+a) being 4x3 a, the 4.3 coefficient of the third term is = 6; annexing the 2 letters with their proper exponents, we have 6 x2 a2 for the third term. In like manner, 6 x2 a2 being the third term, the coefficient of the fourth term is have for the fourth term 4 x3 a. 6.2 3 4; and we Thus, if we multiply the coefficient of any term by the exponent of x in the same term, and divide the product by the number marking the place of that term from the first inclusive, the result will be the coefficient of the succeeding term. ART. 108. Hence, having one term of any power of a binomial, the succeeding term may be found by the following RULE. Multiply the given term by the exponent of x in that term, that is, by the exponent of the leading quantity of the binomial, and divide the product by the number which marks the place of the given term from the first inclusive; diminish the exponent of x by 1, and increase that of a by 1. This rule, which admits of a rigorous demonstration, enables us to develop any power of a binomial. Let it be required to develop (x+a)9. We know, from what has been said, that the first term is 29, and that the second is 9x8 a. The succeeding terms may be found by the rule, and the process of finding them (x+a)9= =x9 +9 x3 a +36 x7 a2+84 x6 a3 + 126 x5 a1 + 126 x4 a5 +84 x3 a6 +36 x2 a7 +9 x a3 + ao. Remembering that odd powers of negative quantities have the sign, we shall find, by means of the rule, that the seventh power of x -a is as follows. (x — a)7 — x7 — 7 x6 a +21 x5 a2. · 35 x1 a3 +35 x3 aa -21 x2a5+7x a6 — a7. · ART. 109. The labor of developing any power of a binomial may be facilitated by attention to the following principles. From the preceding examples and the table of powers given in Article 106, we see, 1. That the number of terms in the development of any power of a binomial, exceeds by 1 the index of that power. Thus the development of (x+a)6 has 7 terms; that of (x — a)9 has 10 terms. 2. If the number of terms in the development is odd, there is one coefficient, in the middle of the series, greater than any of the others; but if the number of terms is even, there are two coefficients, in the middle, of equal value and greater than any of the others. Moreover, those coefficients which precede and those which follow the greatest or greatest two, are the same, but are arranged in an inverse order. Hence, after half or one more than half of the successive coefficients have been found, the rest may be written down without the trouble of calculation. Develop the following quantities. 1. (a+b). 2. (x + y)10. 3. (m−n). 4. (m+n)11. 5. (x — y)12. 6. (x+2y)5. In the 6th example, we must raise the numerical coefficient of y to the requisite powers. We first write the development, merely indicating the powers of 2 y. Thus, (x+2y)5=x5+5x4.2y+10 x3 (2 y)2+10x2 (2 y)3+ 5x (2y)+(2y)5. Now, raising 2y to the powers indicated, and putting the results instead of 2 y, (2 y)2, &c., we have x5+10x4y+40 x3 y2+80 x2 y3 +80 x y1 + 32 y5. 7. (5x+a)1. 8. (a2+x2)5. of a2 To develop the 8th, we first indicate the powers and 22. Thus, (a2+x2)5 = (a2)5 +5 (a2)4 x2 + 10 (a2)3 (x2)2+ 10 (a2)2 (x2)3+5 (a2) (x2)4 + (x2)5. Note, In this example, a2 is the leading term, and the exponents of a2 outside of the parentheses are to be used in finding the coefficients. Now, raising a2 and x2 to the powers indicated, and substituting the results, we have (a2+x2)5= a10 +5 a8 x2 + 10 a6 x2 + 10 a1 x6 +5 a2 x8 +210. As an example in which the binomial theorem may be used to firany power of a quantity consisting of more than two terms, we shall develop (a+b+c). Substitute any single letter, x, for instance, instead of b+c; then a+b+c becomes a +x. Now, (a+x)= a1 + 4 a3 x + 6 a2 x2 + 4 a x3 +x4. But xbc, .. x2=(b+c)2= b2+2bc+c2; 23=(b+c)3=b3+3b2c+3 b c2+c3; and x4 = (b+c)2=b4 + 4 b3 c +6 b2 c2 +4 b c3 + c1. Putting these values instead of x, x2, &c., we have a2+4 a3 (b+c)+6a2 (b2+2 b c + c2) + 4 a (b3 +362 c +3bc2+c3)+64 +4 b3 c + 6 b2 c2 + 4 b c3 + ct. Lastly, performing the multiplication indicated, we have (a+b+c)2 = a1 + 4 a3 b + 4 a3 c + 6 a2 b2 + 12 a2 b c +6 a2 c2 + 4 a b3 + 12 a b2 c + 12 a b c2 + 4 a c3 + b1 +4 b3 c +6 b2 c2 + 4 b c3 + c2. SECTION XXXVI. ART. 110. ROOTS OF MONOMIALS. From the manner in which a monomial is raised to my power, shown in Article 104, results the fol lowing Extract the root of the numerical coefficient, and divide the exponent of each literal factor by the number which marks the degree of the root. The reason of this rule is manifest, as may be shown by an example. Thus, the second power of 3 x y2 is 9 x1 × 2 y2 × 2 = 9x2 y1; consequently, the second root of 9 x2 y1=3x+y=3xy. In like manner, the third root of 27 xo y9 is 3 x* y* = 3 x2 y3. x6 6 It is to be remarked, That every root of an even degree may have either the sign+or -; but a root of an odd degree has the same sign as the power. Thus, the 4th root of a1 may be either a or — a; because (a) = a*, and (-a)+a1. Whereas the third root of a3 is +a, but the third root of - a3 ·a; because (+a)3 =+a3, but (— a)3 = is - a3. We have already said, (Art. 86,) that the second root of a negative quantity is imaginary. The same may be said of any even root of a negative quantity. Thus, the fourth root of 81, and the sixth root of nary quantities. -a are imagi 1. Find the second root of a2 b4. |