see if it will continue to be true when the product is multiplied by a new factor. For this purpose, suppose xm+Axm-1+ Bxm-2+ Сxm-3 . . . +Mam¬n+1+Nxm¬n + + U, ... to be the product of m binomial factors, Nam-n representing the term which has n terms before it, and Mam-n+1 the term which immediately precedes. Let x+k be the new factor by which we multiply; the product when arranged according to the powers of x, will be xm+1+Axm + B xm-1+ C xm-2+ + N xm−n+1 + ... From which we perceive that the law of the exponents is evidently the same. With respect to the co-efficients, we observe, 1st. That the co-efficient of the first term is unity; and 2d. That A + k, or the co-efficient of x", is the sum of the second terms of the m+1 binomials. 3d. Since, by hypothesis, B is the sum of the different products of the second terms of the m binomials, taken two and two, and since Axk expresses the sum of the products of each of the second terms of the m binomials by the new second term k; therefore, B+ Ak is the sum of the different products of the second terms of the m+1 binomials, taken two and two. In general, since N expresses the sum of the products of the second terms of the m binomials, taken n and n, and M the sum of their products, taken n - 1 and n - 1; if we multiply the last set by the new second term k, then N+ Mk, or the co-efficient of the term which has n terms before it, will be equal to the sum of the different products of the second terms of the m + 1 binomials, taken n and n. The last term is equal to the continued product of the second terms of the m + 1 binomials. Therefore, the law of composition, supposed true for a number m of binomial factors, is also true for a number denoted by m + 1. Hence, it is true for m + 2, &c., and is therefore general. 203. Let us now suppose, that in the product resulting from the multiplication of the m binomial factors, we make, we shall then have a = b = c=d: (x + a) (x + b) (x + c) = (x + a)m. The co-efficient of the first term, xm, will become 1. The co-efficient of am-1, being a+b+c+d, . . . will be a taken m times; that is, ma. The co-efficient of xm-2, being ... ab + ac + ad . . . . reduces to a2 + a2 + a2 that is, it becomes a2 taken as many times as there are combinations of m letters, taken two and two, and hence reduces (Art. 201), to The co-efficient of am-3 reduces to the product of a3, multiplied by the number of different combinations of m letters, taken three and three; that is, to In general, let us denote the term, which has n terms before it, by Nam-n. Then, the co-efficient N will denote the sum of the products of the second terms, taken n and n; and when all of the terms are supposed equal, it becomes equal to a multiplied by the number of different combinations that can be made with m letters, taken n and n. Therefore, the co-efficient of the general term (Art. 201), is P(mn+1) an, N= Q x n is called the general term, because by making n = 2, 3, 4, all of the others can be deduced from it. The term which immediately precedes it, is evidently, expresses the number of combinations of m letters taken n and n 1. Hence, we see, that the co-efficient P(m-n+1) P is equal to the co-efficient of the preceding term, multiplied by m Q n+1, the exponent of x in that term, and divided by n, the number of terms preceding the required term. P Since is the co-efficient of the preceding term, we may, by Q observing how the co-efficients are formed from each other, express the co-efficient of the general term thus, The simple law, demonstrated above, enables us to determine the co-efficient of any term from the co-efficient of the preceding term. The co-efficient of any term is formed by multiplying the co-efficient of the preceding term by the exponent of x in that term, and dividing the product by the number of terms which precede the required term. For example, let it be required to develop (x + a). From this law, we have, (x + a) 6 = x6 + 6ax5 + 15a2 x4 + 20a3x3 + 15a2x2 + 6a3x + ao. After having formed the first two terms from the general formula x + maxm−1 +, . . . multiply 6, the co-efficient of the second term, by 5, the exponent of x in that term, and 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)10=x10 + 10ax9 + 45a2x2 + 120a3x7 +210a1x® 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 (x + y) = x2 + 4x3y + 6x2y2 + 4xy3 + y1; and substituting for x and y their values, we have (3a2c2bd)+(3a2c)4 + 4 (3a2c)3 (— 2bd) + 6 (3a2c)2 (— 2bd)2 +4(3a2c) (→ 2bd)3 + (− 2bd)1, or, by performing the operations indicated, The terms of the development are alternately plus and minus, as they should be, since the second term is 205. The powers of any polynomial, may easily be found by the binomial theorem. For example, raise Then (a+b+c)3 = (a + d)3 = a3 +3a2d + 3ad2 + d3; and by substituting for the value of d, (a+b+c)3 = a3 +3a2b+3ab2 + b3 3a2c+3b2c+ 6abc +3ac2 + 3bc2 + c3. This development 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. 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, and perform the operations indicated; then replace the letters introduced by their values. The fourth, fifth, &c. powers of any polynomial can be developed in a similar manner. Consequences of the Binomial Formula. 206. The development of the binomial expression (x+a)m will always contain m +1 terms. Hence, if we take that term of the development which has n terms before it, the number of terms after it will be expressed by m- n. Let us now seek the co-efficient of the term which has n terms after it, and which, consequently, has mn terms before it. We obtain this co-efficient by simply substituting mn for n, in the last value of N in Art. 203. We then have, As we can always take the term which has n terms before it, nearer to the first term than the one which has m n terms before it, we will examine that part of the co-efficient which is derived from the terms lying between these two. We may write m(m-1)... (m-n+1). (m—n). (m—n−1)... (n+2). (n+1) n (n+1). (n+2).. (m-n-1). (m—n)' N= . Now, by cancelling the like factors in the numerator and denominator, we have In the development of any power of a binomial, the co-efficients at equal distances from the two extremes are equal to each other. 207. If we designate by K the co-efficient of the term which has n terms before it, that term will be expressed by Kam-n ; and the corresponding term counted from the last term of the series, will be Kam-non. Now, 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 |