11. Sum the following series by the method of partial fractions:— 12. Find the value of the following to n terms, and ad inf. :— 13. Sum, by reducing to partial fractions, the series given in Ex. (lxix.). 14. Resolve into partial fractions 1 x2 + b2 + c1 − 262x2 - 2c2x2 - 2b2c2* RECURRING SERIES. 177. It has been shown in the former part of this Chapter how fractional expressions can be resolved into partial fractions, and then be developed in a series of ascending powers of the variable x. The inverse problem is, when a series is given, to find the expression from which it is derived; the general term of the series can then be found, also the sum of n terms. Ex. 1. Find the expression of which the series 3 - 7x+20x2 54x3 +148x4, &c., is the expansion. Let a + bx be the expression required; a, b, p, q being 1 + px + qx2 ... a + bx = (1 + px + qx2) (3 7x+20x254x3 +, &c. Or, a + bx = 3+ (3p7) x + (3q − 7p +20) x2 + (20q Equating coefficients of like powers of x, we have 178. Series of this kind are called "Recurring Series," for a reason which will now be given. Let the given series be ao + a1x + α ̧x2 +, &c. + a,x2, &c., ad inf. = be the expression from which the series (1 + px + qx2) (αo + α ̧x + α2x2 + α ̧x3 + Ar xr .) ao + (a1 + pa) x + (α2 + pa1 + αo) x2 + + qar - 2) xr +, &c., ad inf. -1 Here it appears that the coefficients of any three consecutive terms after the second are connected by the invariable relation ar + par •-1+gar-2 = 0. By equating coefficients, as was done in Ex. (1), a。 and a, may be found; then by the relation, a, may be found, then a, and so the coefficients of any number of terms may be found. The denominator 1+ px + qx2 is called the "Scale of Relation." It may consist of 2, 3, 4, or any greater number of terms. Ex. 2. Find two more terms of the series 3 +148x4, &c. 7x+20x254x3 The scale of relation of this series was found in Ex. (1) to be 1+2x+2x2. Hence the coefficients are connected by the relation = 0. ar + par- 1 - 2αr – 2 Ex. 3. Find the general term of the recurring series 1 + 2x + 2x2 - 10x3, &c. By the method of Ex. (1), the expression from which the series is derived is found to be Ex. 4. Find the sum of n terms of the series 1+ 2x + 2x2 The series has been shown in (2) to be equivalent to the two Geo metric Progressions And these, by the for the whole sum 2 (1 + 3x + 32x2 +, &c. ad inf.) (1 + 4x + 42x2 +, &c. ad inf.) formula for the sum of a Geometric series, give Note. If the generating expression can be resolved into partial fractions, the general term and the sum of n terms can be found as in this example. In other cases the process will be more difficult. Ex. 5. Sum the recurring series 5+ 8 +14 + 26 + 50, &c., to n terms. If we take the series 5+ 8x + 14x2 + 26x3 + 50x4, &c., its gene rating expression will be found to be into the two partial fractions 5-7x 1-3x+2x2' which resolves ; and these develop -x and if x be put equal to 1 in each of these, the sum will be ... which is the sum of n terms of the above series. Ex. 6. Find the generating expression of the series x+2x2+5x3 + 12x + 27x5, &c. If we assume the expression to be a + bx 1 + px + qx22 and proceed as in the former examples to equate the coefficients of x on both sides, we shall find the values of p and q obtained from the different equations to be inconsistent with each other. Hence it must be concluded that the expression is not of the form assumed, but must consist of a greater number of terms. If we assume the expression to be a + bx + ca2 1 + px + qx2 + rx3′ sion to be x2 and proceed as before, we shall find the expres (1 − 2x) (1 − x)2° Examples LXXV. 1. Find the expression from which the following series are derived, and write down the general term of the series: 4+9x + 21x2 + 51x3, &c.; 5 + 13x + 35x2 + 97x3, &c. ; 2. Find the generating expression, also the sum of n terms of each of the following: 4+2x+2x2 - 10x3, &c.; 5+ 8x + 14x2 + 26x3 + 50x1, &c.; 7+ 19x+1699x2 + 6259x3, &c.; 4 12x+64x2 240x3 +1024x4032x5, &c. 3. Sum the following series to n terms :— 9 +21 +51 + 129, &c.; - 8+ 4 + 148 + 1376, &c.; 1 + 3 + 11 + 43, &c. ; - 4 + 28 810468; 1 +11 + 89 + 659, &c.; +340 288 +1360, &c. derived; also, that whose nth term is { {3. 2 - 1 − ( − 1 ) ̄`' } 2" - 1. 5. The scale of relation is 1 + ax + bx2, and the 5th and 6th terms are (2a5 - 3a3b - 4ab2) x1 and (2a65a4b 3a2b+3b6)x5, respectively : find the numerator of the generating expression. 6. The scale of relation is 1 term of the series is abr) -1: the series. a+b.x + abx2, and the general find the generating expression of CHAPTER XIX. LOGARITHMS. 179. If some positive quantity different from unity be taken as a base, every number from 0 to x may be regarded as a power of that base. The "Logarithm" of a number is the index of the power to which the base must be raised to produce that number. If 4 be the base, then 42 = 16) 4} = 8) am 102 = 100, or log.10 100 = 2. Since = am- always; if we put m = n; απ Then 1 = a, or log.a 1 = 0 is true for any base. 180. One of the principal uses of logarithms is to shorten the processes of multiplication, division, involution, and evolution, of large numbers. For this purpose 10 is taken as the base, and the logarithms of all numbers are calculated and arranged in tables. The manner of constructing these is beyond the scope of the present work, but the following fundamental propositions are all that are necessary to enable the student to apply the tables to the calculation of products, quotients, powers, and roots. i. The logarithm of a quantity, which is the product of any number of factors, equals the sum of the logarithms of the several factors; or log.a (mnp. .) = log.a m + log.a n + log.a p.... Let a be the base, x, y, z, the logs. of m, n, p, respectively; so that |