2o. The exponent of x in the first term is the same as the number of factors, and goes on decreasing by unity in each of the following terms. 30. The coefficient of the first term is unity. The coefficient of the second term is equal to the sum of the second terms of the binomials; that of the third term is equal to the sum of the different combinations or products of the second terms of the binomials taken two and two; that of the fourth is equal to the sum of the products of the second terms of the binomials taken three and three, and so on. The last term is equal to the product of the second terms of the binomials. 131. We readily infer from analogy, that the same law will obtain, whatever be the number of factors employed. This law may however readily be shown to be general. In order to this, it will be sufficient to show, that if the law be true for the product of any number m of binomials, it will also be true for the product of m+1 binomials. The number of binomial factors being represented by m, the different powers of x will be xm, xm-1, xm−2, &c. Let A, B, C . . . U denote the quantities, by which these powers beginning with xm-1 are to be multiplied; but as the number of terms must remain indeterminate, until m receives a particular value, we can write only a few of the first and last terms of the expression, designating the intermediate terms by a series of points. The product of any number m of factors will then be represented by the expression 3 · 2 + Сxm· U xm + А xm −1 + В xm Multiplying this expression by a new factor x + K, it becomes xm+1+A xm+B xm-1+C xm 2 U K Here the law for the exponents is evidently the same, as in the first expression. With respect to the coefficients, it is evident, 1o. that the coefficient of the first term is unity, 2o A+K, the coefficient of the second term, is equal to the sum of the second terms of the m + 1 binomials. 3o. Since B by hypothesis expresses the sum of the second terms of the m binomials taken two and two, and AK expresses the sum of the second terms of the m binomials multiplied each by the new second term K, B+ A K, the coefficient of the third term, will be the sum of the products two and two of the second terms of the m+1 binomials. In the same manner C + B K, it is easy to see, will be the sum of the products three and three of the second terms of the m + 1 binomials, and so on. 4o. The last term UK, it is evident, is the product of the m+1 second terms. The law laid down, art. 130, being true therefore for expressions of the fourth degree will, from what has just been demonstrated, be true for those of the fifth; and being true for expressions of the fifth degree, it will be true for those of the sixth and so on; thus it is general. 132. If in the different products which have been formed, art. 130, we make b, c and d each equal to a, these products will be converted into powers of x+a, thus (x + α) (x + b) = (x + a)2 = x2 + a | x + a2 a (x+a) (x+b) (x + c)=(x+a)3=x3 + a | x2 + a2 | x2 + a3 a a2 Comparing these expressions with the different products, from which they are derived, we perceive 1o. that the multiplier of x in the second term has been converted into the first power of a, repeated as many times as there are units in the number of binomials employed, or which is the same thing, as there are units in the exponent of x in the first term. That the multiplier of the third term has been converted into the second power of a, repeated as many times, as there can be formed different products from a number of letters equal to the number of binomials employed, taken two and two at a time. 3o. That the multiplier of the fourth term has been con 2o. verted into the third power of a, repeated as many times as there can be formed different products from a number of letters, equal to the number of binomials employed, taken three and three at a time, and so on. 133. From what has been done it is evident therefore, that whatever be the power to which a binomial x + a is to be raised, 1o. the exponent of x in the first term will be equal to the exponent of the power, and that it will go on decreasing by unity in each of the following terms to the last, in which it will be 0. 2o. That the exponent of a in the first term will be O, in the second unity, and that it will go on increasing by unity, until it becomes equal to the exponent of the power to be formed. 3°. The numerical coefficient of x in the first term will be unity, in the second it will be equal to the exponent of x in the first term, in the third it will be equal to the number of products, which may be formed from a number of letters, equal to the exponent of x in the first term, taken two and two at a time, in the fourth it will be equal to the number of products, which may be formed from the same number of letters, taken three and three at a time and so on. Let it be required to form the 5th power of x + a. The different terms, without the numerical coefficients, will be by the preceding rule x5+ a x2+a2x3 + a3 x2 + a1 x + a5 With respect to the numerical coefficients, that of the first term will be 1, that of the second will be 5, that of the third will be equal to the number of products, which may be formed of 5 letters taken 2 and 2, that of the fourth will be equal to the number of products, which may be formed of 5 letters taken 3 and 3, and so on. Thus the numerical coefficients will be 1, 5, 10, 10, 5, 1 whence (x + a)5=x5+5 a x2 + 10 a2 x3 +10 a3 x2 + 5 aa x + a5 Let it next be required to raise xa to the mth power, we shall have, according to the preceding rule, for a few of the first terms without the numerical coefficients xm+a xm-1+ a2 xm−2 + a3 xm−3+ Here the numerical coefficients cannot be determined until we assign a particular value to m; by the preceding rule, however, the numerical coefficient of the second term will be equal to m, whatever the value of m may be. In the development therefore of (x+a)m we write m for the coefficient of the second term. With respect to the third term the numerical coefficient will be equal to the number of products, which may be formed of m letters 2 and 2 at a time? this is expressm(m-1) ed by the formula we write therefore 1.2 the coefficient of the third term. For a similar reason m (m-1) (m2) 1. 2. 3 m(m—1) for will be the coefficient of the fourth term and so on. We have then m(m-1) 1.2 (x+a)m—xm+maxm−1+ -a2xm-2+ From inspection of the different terms of this development it will be perceived, that the coefficient of the fourth, for ex ample, is formed by multiplying third term, by m - 2 the exponent of x in this term, and dividing by 3 the number, which marks the place of this term. It will be perceived also that the coefficient of the third term is formed in the same manner by means of the second term, and that of the second by means of the first. We readily infer therefore the following rule, by which to form the coeffi cient of any term whatever, viz. Multiply the coefficient of the preceding term by the exponent of x in that term, and divide the product by the number, which marks the place of that term from the left. From what has been done, we have therefore the following rule, by which to raise a binomial to any power whatever, viz. 1o. The coefficient of x in the first term is unity, and its exponent is equal to the number of units in the degree of the power to which the binomial is to be raised. 2°. To pass from any term to the following, we multiply the numerical coefficient by the exponent of x in that term, divide by the number which marks the place of that term from the left, increase by unity the exponent of a and diminish by unity the exponent of x. According to this rule (x+a )=x6+6 a x5+ 15 a2x2+20 a3x3 + 15 a2x2 +6 a3x+a¤ 134. It sometimes happens, that the terms of the proposed binomial are affected with coefficients and exponents. The following example will exhibit the course to be pursued in cases of this kind. Let it be proposed to raise the binomial 4 a2 b-3 a b c to the fourth power. Putting 4 a2 b=x, and 3 a b c = Fy, we have (x + y) = x2+4x3 y +6 x2 y2 + 4 x y3 + y2 Substituting next for x and y their values, we have 3 (4 a2b-3 a b c) (4 a2 b)4 + 4 (4 a2 b) 3 (-3 a bc)+.. 6 (4 a2 b)2 (— 3 a b c)2 + 4 (4 a2 b) (— 3 a b c) 3 + (— 3 a b e) 4 or performing the operations indicated, we have (4 a2 b— 3 a b c) 4 = 256 a3 b1 — 768 a7 ba c + 864 a6 b4 c2 . . − 432 a5 b1 c3 + 81 aa b1 c1 The terms produced by this development are alternately positive and negative. This, it is evident, should always be the case, when the second term of the proposed binomial has the sign 135. The powers of any polynomial whatever may be found by the binomial theorem. Let it be proposed to find for example, the third power of the trinomial a+b+c. m; In order to apply the rule to this case, we put a+b= the proposed is then reduced to the binomial m+c, and we have (m + c)3=m3 +3 m2 c +3 m c2 + c3 whence, restoring the value of m, we have (a+b+c)3 = a3 + 3 a2 b + 3 a b2 + b3 3 a c2 + 3b c2 + c3 The same process, it is easy to see, may be applied to any polynomial whatever. MISCELLANEOUS EXAMPLES. 1. To find the third power of 2 a b + c2. 2. To find the seventh power of 3 a5 — 2 a2. |