Substituting these values of B, C, D, &c., in equation (2,) we shall have Any Algebraic expression consisting of two terms is called a binomial; as a +x, c + b, &c. We propose to find a rule to raise a binomial to any power, without the trouble of continued multiplication; for this purpose it is necessary to see how the first and second terms of a binomial are formed. (a+x): = a*+4 a3 x + 6 a2x2 + 4 a x3 +xa It is easy to see that the first term in the power, is the first term of the binomial, raised to the same power to which the whole binomial is raised. The first term in the second power of a +x is a2, the first term in the third power of a + x, is a3, &c. Now, observe how the second term, or the term containing the first power of x, is formed. In the second power of ax, the second term is 2 ax. In raising the binomial to the third power, this term is multiplied by a, which raises it to 2 a2 x. We get another term, similar to it, by multiplying a2 of the second power by x; these make 3d x. This is the second term of the third power. When we multiply 3 ax by a, in raising the binomial to the fourth power, it becomes 3 a3 x. We obtain a similar term by multiplying the a3 by x; which gives a3 x. These two united give 4 a3 x; which is the second term of the fourth power. It is evident that in every succeeding power, the exponent of a increases 1, and the coefficient of the term increases 1. The second term in the fifth power will be 5 ax; of the sixth, 6 a3 x; and, the second term of the n power na”—1 x. One term of the binomial may be a compound term, since two terms may be considered and used as one. (a + x)+ h may be considered as a binomial, calling a +x one term. Then, if (a + x)+h be raised to the second power, the second term will be 2 (a+x) h, the second term of the third power of (ax)+h will be 3 (a + x)2 h, the second term of the n power is n (a + x)-1 h, &c. The second term of three, four or five times any power of (a+x) + h is three, four or five times as much as the second term of the same power of (a + x)+ h; thus the second term of 6 6 ((a + x) + h)" ; is 6n (a+x)-1 h, &c. Now, it is required to find a series of single terms equivalent to (a + x)". Assume (A) (a + x)" = A + B x + Cx2+Dx2+ Ex1, &c. This is an analytical equation, and must be true, “whatever x may be; and, consequently, it is true when x = 0, or when x=x+h. Now, if we substitute 0 in place of x in equation (A,) we have in the first case If we substitute x+h for x in equation (A,) we have (C) ((a+x) + h)" = A + B (x + h) + C (x + h)2 + D (x +h,) &c. Finding the equivalent single terms of the two members of this equation, we have the second term, or the term multiplied by the first power of h, in the first member, equal to `n (a+ 2) h. This is the only term multiplied by the first power of h, in the first member; and, consequently, it must be equal to all the terms that are multiplied by h in the second member, These terms in the second member, are Bh+2Cxh+ 3D x2 h, &c. Hence n (ax)1h=Bh+2Cx h + 3D x2h, &c. (D) n(a+x)=B+2Cx+3Dx2, &c. This last equation is analytical, and must be true when x = 0, or when x=x+h, Substituting these two values successively, we have in the first case (E) and, in the second n na"-1 = B · ((a+x) h) (a + x) + h ) 11 = B + 2 C (x + h) + 3 D (x + h)2, &c. Finding the second terms, or the terms multiplied by the first powers of h in both members, and putting them equal, we have n(n-1) (a + x)-2 h = 2 Ch+6Dxh, &c. Dividing by h, and making x = 0, we have Substitute the values of A, B, C, D, &c., found in equations (B,) (E,) (F,) (G,) in place of A, B, C, D, in equation (A,) and we have (H) (a + x)" = a"+na+x+ n (n − 1) an—2x2 -2 + 2 2. If n 1, what will equation (H) become? If n = 2? 3? 4? 5? 6? 7? 8? 9? 10? 11? 3. What is the fourth power of b+c? Fifth power of b+c? 4. What is the tenth power of a+b? The 11th? 16th! By the same process by which we found the value (a+x)", we find 5. What is the seventh power of (a—b)? Of (b—c)? SECTION CIII. The Method of Differences. It is generally easy to find a series equivalent to any compound term, or finite expression. It is not so easy to find a compound term, or finite expression, equivalent to a given series. We have already seen how to find any given term, or the sum of any given number of terms of an arithmetical or geometrical series, when the first term and common difference, or first term and quotient, are given. The method of differences is applicable to many other kinds of series. Take the second powers of the numbers 1, 2, 3, 4, 5, 6, 7, &c., and we have 1, 4, 9, 16, 25, 36, 49, &c. Take the differences between the first and second, second and third, third and fourth, fourth and fifth terms, &c., and these differences make a series called the first order of differences. 'Take the differences of the consecutive terms of this last series, and we have another series called the second order of differences, &c.: thus, 1, 4, 9, 16, 25, 36, 49, &c. 3, 5, 7, 9, 11, 13, First order of differences. 2, 2, 2, 2, 2, Second order of differences. In order to see the general law, let a, b, c, d, e, f, g, h, &c. be a series. ba, c—b, d—c, e— -d, fe, &c. The 1st order of diff. +a-2b+c, +b−2c+d, &c. The 2nd order of diff. -a+3b-3c+d, &c. The 3rd order of diff. It will be seen, that the letters in the first terms of the several orders of differences, have precisely the same coefficients, that the terms in the corresponding powers of the binomial a-x have; except that in the odd orders, as the third, fifth, &c., the terms have contrary signs. Hence, if we wish to find the first term of the second order of the series a, b, c, d, &c., we have it a-2b+c, the first term of the third order a+3b-3c+d, and for the first 1. Required the first term, of the third order of differences, of the following series: 2, 6, 12, 20, 30, &c. It will be by substituting the numbers in place of the letters a, b, c, n, &c. in the series (B) -2+3x6-3X2X12 + 3X2X1X20 = 0 Hence the first term of the third order is = 0. 2. Find the first term, of the third order of differences, of the series 1, 8, 27, 64, 125, &c. 3. Find the first term, of the fourth order of differences. of the series Suppose it is required to find any given term of a series, Let a, b, c, d, &c. be the series, and m', m", m''', m'''', &c. the first terms of the first, second, third and fourth order of differences. |