OF SERIES. A SERIES is an assemblage of terms, which continually increase or decrease according to a certain law, as the arithmetical and geometrical series spoken of before. A Converging Series is that of which the terms continually decrease, and a Diverging Series is one of which the terms continually increase. Series are obtained by division, by the extraction of roots, and by various other operations. 2.4.6a5 2-4-6-8a7 LEMMA. If the series (a+b)x+(c+d) x2+(e+f)x3, &c. continued indefinitely, be always nothing, whatever be the value of x, then the coefficient of any one power of x is 0; that is, a+b=0, c+d=0, &c. For if the equation be divided by x, then a+b+(c+d)x+(e+f)x2=0. Here let x = 0, then a+b=0; therefore (c+d)x+(e+f)x2 =0, whatever be the value of x; and proceeding in the same way we find c+d=0, and so on. A GENERAL METHOD OF FORMING SERIES. Assume a series with unknown but constant coefficients, and having the indices of x increasing or decreasing, in the same way as if the operation were performed at large; then make this series equal to the given quantity, and having cleared the equation of surds and fractions, bring all the terms to one side, so as to make the equation = 0; after which make the sum of the coefficients of each power of x=0, which will give as many equations as there are unknown coefficients; and therefore the values of these coefficients may be found, and substituted for them in the assumed equation. 1. Required a series = a 6+x Dx3, &c. =; and by multiplying by b+a, and transposing, we get Ab−a+(Bb+A)x+(Cb+B) x2, &c. = 0, an equation which must be true, whatever be the value of x. Therefore Ab- a=0, Bb+A=0, Cb+B=0, &c. whence a -a +a A=+, B== C=, &c.; and these values, substi 639 ax 3 649 &c. tuted for A, B, C, &c. in the assumed equation, give ax 2 + b3 a +x Ans. 1+3x+4x2+7x5 +11x4+18x5, &c. This is a recurring series, in which each of the coefficients after the second is the sum of the two preceding ones. SUMMATION OF SERIES. To sum a series is to find a terminated expression equal to the interminate series. I. Let a+b+c+d, &c. be any series; subtract each of the terms from the one following it, and the differences will be −a+b, —b+c, −c+d, &c. This is called the first order of differences. Subtract each of these from the following for a second order of differences, viz. a-2b+c, b−2c+ď, -2d+e, &c. Subtract these again to get another order of differences, and so on. II. Let d' be the first term of the first order of differences, d" the first term of the second order, and d'", d'""', d''''', &c. the first terms of the following orders; then d'=—a+b, d" a -2b+c, d'"'=a+3b 3c+d, d"""= a — 4b +6c-4d+e, &c., from which we infer that da n- -1 n -2 nb±n. ·c In. 21. "2d, &c. to n+1 terms; in which series the upper signs are to be used when n is an even number, and the under when it is odd. Required the first of the fifth differences of the series 6, 9, 17, 35, 63, 99, 148, &c. Required the first of the sixth order of differences of the series 3, 6, 11, 17, 24, 36, 50, 72, &c. Ans. - 14. III. Again, since d' a+b, therefore ba+d'; and in the same manner we get ca+2d'+d", d=a+3d'+ 3d"+d"", e=a+4d′+6d" +4d""'+d'""', &c., and therefore * Here n is not the exponent of a power, but the index of the order of differences. the nth term of the series = a+(n−1)d'+”—1."—2d" + Required the 7th term of the series 3, 5, 8, 12, 17, &c. Here d'2, d'1, d""=0, and the 7th term, or z= 3+6·2+6·2·1=3+12+15 = 30. 2 What is the 9th term of the series 1, 5, 15, 35, 70, &c.? Ans. 495. IV. Also, if these values of a, b, c, &c. be added, we obtain 2a+d'=a+b, a+b+c=3a+3d'+d", a+b+c+d= 4a+6d'+4d"+d"", &c. Whence we conclude, that the sum of n terms s=na+n' 2 n-1 n-2 ・d'+n⋅ n=2d", &c. 2 3 If the differences come at last to be equal, these two last series will terminate, otherwise they will be interminate. 1. Required the sum of 8 terms of the series 2, 5, 10, 17, &c. 7 Here d′ = 3, d'′′ = 2, d′′"=0; therefore s=8·2+8· ·3+ 8-12=16+84 +112=212. 76 23 2. What is the sum of 12 terms of the series 21, 56, 126, 252, 462, 792, &c.? Ans. 27125. 3. Required an expression for the sum of n terms of the fourth order of figurate numbers, 1, 4, 10, 20, 35, &c. Here d'=3, d"=3, d"" =1, and d'""'0, and s=n+ = ··3+n⋅ 3 ··3+n⋅ n+1 2 .1, which, 4 Thus 12 terms are 2 3 =1365. The number of factors in the formula, and the order of differences which become = 0, are the same with the order of the figurates. 4. What is the sum of n terms of the squares (m±a)2+ (m+2a)2+(m±3a)2, &c. + (m+na)2? 5. Required the sum of 12 terms of the series 3o +5o+ 72 +92, &c. Here m = 1, a = 2, and n = 12; therefore the sum tuted for A, B, C, &c. in the assumed equation, give ax 3 b2 + &c. 1 ―X- x2 a Ans. 1+3x+4x2+7x+114+18x5, &c. This is a recurring series, in which each of the coefficients after the second is the sum of the two preceding ones. SUMMATION OF SERIES. To sum a series is to find a terminated expression equal to the interminate series. I. Let a+b+c+d, &c. be any series; subtract each of the terms from the one following it, and the differences will be ·a+b, —b+c, -c+d, &c. This is called the first order of differences. Subtract each of these from the following for a second order of differences, viz. a-2b+c, b—2c+d, c-2d+e, &c. Subtract these again to get another order of differences, and so on. &c. - a+b, .46 II. Let d' be the first term of the first order of differences, d" the first term of the second order, and d'", d'""', d''''', the first terms of the following orders; then d'—— d"—a—2b+c, d'"'=—a+3b—3c+d, d'""' = a— +6c-4d+e, &c., from which we infer that d* =±a‡ n- -1 n=1 n=2d, &c. to n+1 terms; nb±n. c = n which series the upper signs are to be used when ʼn is an even number, and the under when it is odd. Required the first of the fifth differences of the series 6, 9, 17, 35, 63, 99, 148, &c. =+3. Required the first of the sixth order of differences of the series 3, 6, 11, 17, 24, 36, 50, 72, &c. III. Again, since d'――a+b, therefore b= in the same manner we get ca+2d'+d", d=a+3d' + 3d"+d"", e=a+4d'+6d" +4d""'+d'''', &c., and therefore * Here n is not the exponent of a power, but the index of the order of differences. the nth term of the series = a+(n−1)d'+”—1."—2 d′′+ Required the 7th term of the series 3, 5, 8, 12, 17, &c. Here d'2, d"= 1, d""=0, and the 7th term, or = 3+6·2+6·2·1=3+12+15=30. 2 What is the 9th term of the series 1, 5, 15, 35, 70, &c.? IV. Also, if these values of a, b, c, &c. be added, we obtain 2a+d' = a+b, a+b+c=3a +3d'+d", a+b+c+d= : 4a+6d′+4d"'+d"", &c. Whence we conclude, that the sum n-1 n-2 of n terms s=na+n·· -d'+n⋅ 3 n-1 2 -2d", &c. If the differences come at last to be equal, these two last series will terminate, otherwise they will be interminate. 1. Required the sum of 8 terms of the series 2, 5, 10, 17, &c. Here d′ = 3, d'' = 2, d'"'=0; therefore s=8·2+812·3+ 8216+ 84+112=212. 76 23 2. What is the sum of 12 terms of the series 21, 56, 126, 252, 462, 792, &c.? Ans. 27125. 3. Required an expression for the sum of n terms of the fourth order of figurate numbers, 1, 4, 10, 20, 35, &c. Here d' 3, d"=3, d"" =1, and d'""'0, and s=n+ n-1 n 2 n- 3 n 1 2 = •1, which, n- - 1 N -2 n. ··3+n·· The number of factors in the formula, and the order of differences which become = 0, are the same with the order of the figurates. 4. What is the sum of n terms of the squares (m±a)2+ (m+2a)2+(m±3a)2, &c. + (m±na)2? Ans. nm2n." • 2ma+n.• " + 1 2n+1 n+1 2 •a2. 5. Required the sum of 12 terms of the series 32+52 + 72 +92, &c. Here m = 1, a2, and n = 12; therefore the sum |