+&c. Here, leaving out the numerators, and the last factor of each denominator, let there be assumed 1 1 m+rm+2r + + + +&c. to n terms + m+r m+2rm+3r Whence, by subtraction and substitution, r r m(m+r) + (m+r) (m+2r) + (m+2r) (m+3r) ' 1 +&c. to n terms Where, if the number of terms (n) be increased 1 mr + nr2 Or, which is the same, will vanish, and the sum of the series the reciprocal of the product of the two quantities m and r. In which case, if m and r be each taken = 1, we shall have 1.2 +2.3 +3.4+4.5+ &c. to n terms And by taking m and r of different values, the sum of n terms of a variety of other series may be found. 139. Required the sum of the infinite series rators, and the last factor of each denominator, as before, let there be assumed m(m+r) (m+r)(m+2r)' (m+2r)(m+3r) (m+r) (m+2r) *(m+2r)(m +3r) †(m+3r) (m+4r) m(m +r)(m+2r)' (m+r)(m+2r)(m+3r) 2r 1 2rm (m + 2r) In which case, the sum of n terms of the series, determined as in the 139th will be = 1 2(n+1)(n+2) +3.4.5+&c. to n terms, = 1 4 And by taking m and r of different values, the sums of various other series of this kind may be found, as in the preceding example. Questions without Solutions. 140. Sum 100 terms of the series 2, 5, 8, 11, 14, &c. and 50 terms of the series 1+2+3+4+52+&c. Answers, 15050, and 42925. 141. Required the sum of the infinite series 1+3x+6x2+10x3 4-15x+&c. when x is less than 1. 142. Required the sum of the infinite series 1+4x+102+20x3 +35x+&c. when x is less than 1. &c. 1 1 1 1 1 143. Sum the infinite series + + + + + 1.3 3.5 5.7 7.9 9.11' 144. Sum 40 terms of the series (1X2) + (2×3) + (3×4) + (4×5)+ &c. 145. Sum n terms of the series n terms of the series 1+8x+27x+64x3+125x1, + &c. 2x -5 2x 2x +2.3.4.5 +&c. and 147. Required the sum of n terms of the series (r—1)2 1, nr+r-n of an infinite number of terms, and s the sum of n terms. 148. Required the sum of the series +2n(4 + 2n)* The symbol, in this and the following series, denotes the sum 1 2.6 +4.8 + 12+6n27+9n +6.10 1 n +&c... 3 5n+3n2 16' 32+48n+16n2 1 3.8 +6.12 +9.16 +12.20 Ans. Σ 150. Required the sum of the series + + + 7.12 12.17 3n 2+5n 1 6.10 8.12 10.14 n 16+16n 36+24n 16+16n36+24n +14.27+&c. + n 12+ 12n 6 2.7 3 Ans. S= 1 1 + - 1 n 48' 12.24 +30(10+2n)* 1 12.12 15.14 n 1 ±3n(4+2n)* 154. Sum n terms of the series 2(3+6n) 4(6+6n) 1 1.2.4 +2.3.5 +3.4.6 n 12(2+n) 2 3 4 3.5 +57 +5.7-7.9 Ans. : 12' S= 1 1 1 4.5.7 n 18(3+n)' 5 9.11+ &c. 1 —124(3 + 4n) * 1 4 5 + &c.+3n(2n−1).(2n+1)* 1 Ans. = 1 44.3n(1+2n) 161. Let ab3x2+(1+c)bd/c+cb2x2={b3d✔c+(ab+c)(1+c)}x. to find the values of x. Let x = By tr. b2(ab+c)x2 —{b3dc+(ab+c)(1+c)} x =—(1+c)bdc. 22—{b3d√c+(ab+c)(1+c)} z=—b3d(at+ ́b2 (ab + c)' c)(1+c)c. Let mb3d/c, and n = (ab+c)(1+c), then will z2 (mn) z=mn, by art. 70, case 4, I have z=√√√ { } (m2 —2mn +n2)}±{(m+n); b2(ab+c)x, or z=m, or n; b3dc bdc b2 (ab+c) ab+c; and x = (ab+c) (1 + c ) 162. A merchant having mixed a certain number of gallons of brandy and water, found, that if he had mixed 6 gallons more of each, there would have been 7 gallons of brandy to every 6 gallons of water; but if he had mixed 6 gallons less of each, there would have been 6 gallons of brandy to every 5 gallons of water: how many of each did he mix? Let 7x-6, and 6x-6 denote the number of gallons of brandy and water respectively; which assumption fulfils the first condition of the question. By the second condition we have 7x—12 : 6x— 12::6:5, or x : 6x—12;;1 : 5, whence 5x=6x-12, or x=12; and 7x-6 78 gallons of brandy, 6x-6= 66 gallons of water. 163. Find two numbers, such, that their sum, product, and difference of their squares, shall be all equal. The sum of two numbers differing by unity, is equal to the difference of their squares. Let, then, x+1 and x denote the numbers then their sum must be equal to their product; that is, 2x+1=x+x, or x-x-1 by case 2; 4x-4x+1=4+1=5 extracting the root, 2x-1=5, .. x=}±}No5; and x+1= 5, the two numbers required. 164. Find two numbers, such, that their product may be equal to the difference of their squares, and the sum of their squares equal to the difference of their cubes. If we assume ax and (a+1)x for the numbers, a being +5, we shall fulfil the first condition of the question. (See Quest 163.) By the second condition, we must have (2a2+2a+1)x2=(3a2+3a+1)3; or (since a2 a +1). Hence, ax = 2a2+2a+14a+3 3a+3a+1 6a+4' 4a2+3a7a+4_4+3No5 6a+4 6a+4 7+35 and (a + 1)x = a2x = ax × a ={√5(}+{√5)=4(5+No5) : · that is, the numbers are 5, and 1(5+5). 165. Given (6x+8) — † (5x+3)=§ (27 — 4x) — § (3x +9), 27-9x 5x+2_61 2x+5 29+4x 11 166. Given x+ the value of x. 7x-8, 15x+8 to find 12 167. Given + =3x 13 to find x. x= 9. x = 3. 2 169. Given 9b2-3a2b2 X(a+b)c, to find the values of x. (5No (a+b), (1+262)cdo 168. Given (5x— 1)— 4(7x— 2)—63—§x, to find x. 5a+10ab2 (5√(a+b) (1+262)cdc cd 5a(1+262) By tr. 362 (3-a2) + 363 cd/cNo (a+b) 1+262 Let m ; and n= ab 3 a2 have zn+mcdc)}2 = {(n+mcdc)2 — cdmn c; or {z—(n+mcd/c)}2={{n2 — { mncd/c+4(m2c2d°c); root = x — (n+mcdc)=(n-mcdc). Hence I have |