391. To apply these formulas in the summation of any given series, we must first determine the scale of relation by (P) or (T). and then we may obtain the sum of the series from (@) or (V). If the order of the series is not known, we should first determine the values of m and n by formula (P), and ascertain by trial whether the scale of relation thus found will apply to the given series. If it will not apply, we may determine the values of m, n, and r from formula (T), and ascertain by trial whether the series can be developed by the new scale thus obtained. If this also fail, we must establish other formulas corresponding to still higher degrees, and continue the trials. If, however, we resort in the first place to a formula corresponding to an order higher than that of the given series, then one or more of the quantities, m, n, r, etc., will prove to be zero, and the remaining numbers may be taken as the scale of relation, without further trial. 1. Find the sum of the infinite series, 1 + 4x + 10x' + 2230° + 46x*+.... To determine the scale of relation, we have a=1,1 = 4, c= 10, d = 22. These values substituted in formula (P), give m+4n = 10, 4m+10n = from which we readily obtain -2, These numbers form the true scale of relation ; for we perceive that any coefficient after the second in the given series, is equal to three times the first preceding coefficient, minus twice the second preceding coefficient. To find the sum of the series, we have B Whence by formula (Q) 1+4x—32 s Ans. 1-3x+2x* 1-3x+2x?' We have thus obtained the sum of the series in the form of an 22, m = n = 3. A = 1, = 4x. 1+2 algebraic fraction. Conversely, the given series may be developed from this fraction, either by division, or by the method of indeterminate coefficients. Indeed, it will be found that the sum of every recurring series is an irreducible fraction, from which the series may be supposed to originate. The fraction from which any particular series is supposed to arise, is called the generating fraction for that series ; it is the same as the sum of the series. EXAMPLES FOR PRACTICE. 1. Find the sum of 1+3x+4x+7x+11x*+.... 1+22 Ans. 1-x-3 2. Find the sum of 1+6x+12° +48 +120x*+.... 1+52 Ans. 1-2-6xca 3. Find the sum of 1+2x–5x+26x'--119x*+.... 1 + 6% Ans. 1+ 4x 2.;** 4. Find the sum of 1 + 4x + 3x— 2x' + 4x* + 17.x® + 3x +.... 1+3x+2" A18. 1-x+ 2x - 30 5. Find the sum of 1+3x+5x+7+9x*+.... 1+ Ans. (1-x) 6. Find the sum of 1+:+5x+13x* +413* +121x*+. 1-X Ans. 1-2x-3x 7. Find the sum of 1+4x+6x* +11x +28x* +63x +... (1+x)'—2x Ans. (1-x) 3.2 7x2 61.7 8. Find the sum of + 10x'+ + sc*+ +913" +... 2 2 2 Ans. 2-406x** DIFFERENTIAL METHOD. 392. The Differential Method is the process of finding any term of a regular series, or the sum of any number of terms, by means of the successive differences of the terms. 393. To find any term of a series by the differential method. If we subtract each term of a series from the next succeeding term, we shall obtain a new series called the first order of differences. If we subtract each term of this new series from the succeeding term, we shall obtain a series called the second order of differences; and so on. Let a, b, c, d, e, .. represent a regular series, the successive terms being formed according to any fixed law. We will write the given terms in a vertical column, and proceed by actual subtraction to form the several orders of differences, placing each order in a separate column, and each difference at the right of the subtrahend. The result is as follows : The quantity which stands first in any column, though a polynomial, is called the first term of the order of differences which designates the column. Let d,, d2, dz, dq, etc., represent the first terms of the first, second, third, fourth, etc., orders of differences. Then we shall have di=b-a By transposition, we may obtain, after making the necessary substitutions, a = a b=a+d e=a+4, +6d2+4d2 +d4, etc. These equations express the values of a, b, c, d, e, etc., in terms of a, dj, dz, dz, dq, etc. The coefficients in the second members are formed according to the binomial formula ; and we observe that the coefficients of the second power of a binomial are found in the third equation, the coefficients of the third power in the fourth equation, and so on. Hence, if we let Toti denote the (n+1)th term of the given series, a, b, c, d, eg.... then we shall have n(n-1) n(n-1)(142) T.+1 = a+md + dat det.... (m) 2 2 3 And by substituting n–1 for n, we shall obtain a formula for the nth term of the given series ; thus, (n-1)(n-2) (n-1)(n-2)(n-3) at(n-1)d, + dat (A) 2 2 3 394. To find the sum of any number of terms of a series, by the differential method. Represent the given series by a, b, c, d, e,.... (1) And denote the sum of n terms by S. We are to find the values of S in functions of a, da, da, dz, etc. Let us assume the auxiliary series, 0, a, a+b, a+b+c, a+b+c+d, (2) It is obvious that the (n+1)th term of this series is the same as the sum of n terms of the given series, (1), and may be placed equal to S. Now let d'y d' 22 d'a, etc., represent the first terms of the successive orders of differences in the auxiliary series (2). Then by formula (A), we have n(n-1) (1-2) 0 d'z to... (n) 2 3 But if we proceed to form the first, second, third, etc., orders of differences for the auxiliary series (2), we shall have d'1 = a, b-a = d1, d': = C--26+a= d.,, etc. Hence, by substitution in equation (n) we have n(n-1) n(n-1)(n —2) d2+.... (B) 2 2 12, 395. The use of formulas (A) and (B) may be illustrated by the following examples : 1. Find the 12th term of the series, 1, 5, 15, 35, 70, 126, etc. We first form the successive orders of differences, as follows: 1, 6, 126, 56, 21, 6, 1, 0. Thus we have n = and a=1, d=4, d. = 6, dz=4, d=1, do=0. Substituting these values in (A), and reducing the terms, we obtain 112=1+44+330+660+330 =1365, Ans. The series is broken off at the fifth term, because the subsequent differences are all zero. 2. Sum the series 1, 3, 6, 10, 15, 21, etc., to n terms. By forming the successive orders of differences, as in the last example, we shall obtain 1, d, = 2, d2 = 1, dz = 0. Whence, by formula (B), a |