(556.) It is evident that this method may be very easily extended to any series whose general term is of the form 1 (mx+a){m(x+r)+a}{m(x+2r)+a}...{m(x+n—1r)+a} If the terms of the series have a constant numerator, as p instead of 1, the sum will just be p times greater. By methods somewhat similar, the sum of any series, the numerators of whose general term is of the form px + e. and denominator of the above form, may be found, when the denominators consist of more than two terms. (557.) Prob. V. To find the sum of the series whose general term is Since px+e (mx+a){m (x+r) +a} {m (x+2r)+a} m (px + e) (mx + a) { m (x+r) + a} { m (x+2r) + a} (mx+a) {m (x+r) + a} {m (x+2r)+a} If S denote the difference between the sums of the two series whose general terms form the second member of this equation, then the sum of the proposed series is proved, as Find the sum of the series whose general term is x+2 (2x+1)(2x+3) (2x+5) = Herep 1, e2, m = 2, a=1,mr=3—1=2; hence r=1, and p (2 mr + a) — me =1 (2x+1)(2x+3) (2 x + 1) (2 x + 3) (2x+5) Find the sum of the series whose general term is (558.) Prob. VI. To find the sum of a series whose general term is px + e (mx+a){m(x+r)+a} {m (x+2r)+a} {m (x+3r)+a} It may be shown, as in the last problem, that the sum of the proposed series is the mth part of the difference between two series whose general terms are and p (mx+a) {m (x + r) + a} {m (x+2r)+a} p (3 mr + a) -me (mx+a){m (x+r)+a}{m(x+2r)+a}{m(x+3r)+a} EXAMPLE. Find the sum of the series whose general term is 2x+3 x(x+1)(x+2)(x+3) Here p = 2, e = = 3, a = 0, m = 1, rm = 1; hence r― 1, and p (3r+a) — me = 3 EXERCISE. Find the sum of the series whose general term is 4x+6 x(x+1)(x+2) (x+3) (559.) The sum of a series, such as S .1218 Sum= 3 px2 + qx + c (x+a)(x+b)(x+c)' may be found by a method somewhat similar. Thus, assume = Ax+B px2+qx + c C (x+a)(x+b)(x+c) ̄(x+a)(x+b) (x+a)(x+b)(x+c) and by reducing the fractions in the second member to one, and equating its numerator with pa2 + qx+c, the values of A, B, and C, would be found by the principle of undetermined coefficients, as their values remain constant, while those of a vary; and if b-ac-b, then by the preceding methods, the sums of the series of which the terms in the second member are the general terms, can be found, the difference between which would be the sum of the proposed series. This method may be easily extended. REVERSION OF SERIES. (560.) When the value of one quantity is expressed in terms of another by an ascending series, the value of the latter may also be similarly developed by the method of reversion of series, called also the inverse method of series. (561.) Let y = ax + bx2 +cx3 + dxa + [1] the coefficients a, b, c, being known; in order to find the developement of x in terms of y, assume the series x = Ay+By2+ Cy3 + Dya +... ... ... [2] in which the coefficients A, B, C, are undetermined. Find the values of y2, y3, ya, ... + 2 ac a3x3+3a2bx ... from [1], thus:— + ... Substituting these values in [2], and arranging +3 Ca2b + Da4 24 + ... Equating with zero the coefficients of the different powers of a (469), Aa-10, Ab+ Ba2 = 0, Ac+2 Bab + Ca3 = 0, Ad + Bd2+2 Bac +3 Ca2b+ Da1 = 0, ... from which are derived Hence the developement of x in terms, of y is by [2] a 1 b ac- -262 a2d-5abc+5b3 a5 a7 (562.) If the given series has a constant term prefixed, thus, ... [1] [2] y= a + ax + bx2 + cx3 + dxa + then assume y ... and the value of x developed in terms of z is found in the same manner as its expansion in terms of y was found before (561), by assuming x= Az+Bz2 + Cz3 + Dz1 +. the coefficients of which would be found to have the same value as before; so that if z be substituted for y in [3], the result is the required developement of x; and then y being substituted for z, the result is (563.) When the given series contains the odd powers of x, assume for x another series containing the odd of y. Thus, let y = ax + bx2 + cx5 + dx2 + ... to develope a in terms of y, assume x = Ay + By3 + Cy3 + Dy1 + ... powers Substitute in the latter series the values of y, y3, y5, derived from the former in the same manner as was done in (561); then, having arranged the result according to the powers of x, the coefficients of x, x3, x5, ... being each equated with zero, the values are found as formerly to be The following exercises will be easily solved by substituting for a, b, c, d, ... their values in the given series, and then finding the values of A, B, C, D, ... as in the following example: 1 Let y=1+x+= x2+ 2 ·203+ 1 find the developement of x in terms of y. By [1] in art. (562), it appears that here a= 1, a = 1, A, B, C,... which are the same as in art. (561), are |