Ans. 1+3x+4x2+7x3+11x1+18x5+, &c. This is a recurring series, in which each of the coefficients, after the second, is the sum of the two preceding ones. 5. Expand (1-a) into a series. Ans. 1+x+5x2+13x2+41x+121x+365x, &c. 7. What is the expansion of (a—b)‡ ? Ans. a at (1– 4a 4.8a2 4.8.12a3 4.8.12.16a1 8. It is required to expand (a+x)—2. 1 2x 3x2 4x3 Ans. + a3 at +, &c. 5 SECTION XXXVI. SUMMATION AND INTERPOLATION OF SERIES. ART. 334. The Summation of Series is the method of finding a terminated expression equal to the whole series. Interpolation is the method of finding any term of an infinite series, without producing all the rest. DIFFERENTIAL METHOD. 335. The Differential Method consists in finding, from the successive differences of the terms of a series, any intermediate term, or the sum of the whole series. PROBLEM I. 336. To find the several orders of differences. Let a+b+c+d+e+, &c., be any series; subtract each term from the one following it, and the differences a+b, −b+c, -c+d, −d+e, &c., will form a new series, called the first order of differences. Again, subtract each term of this new series from the one that follows it, and the differences a-2b+c, b-2c+d, c-2d+e, &c., will form another series, called the second order of differences. Proceed in like manner for the third, fourth, fifth, &c., order of differences, until they at last become equal to 0, or are carried as far as is required. 337. When the several terms of the series continually increase, the differences will all be positive; but, when they decrease, the differences will be alternately negative and positive. 1. Required the several order of differences of the series 1, 6, 20, 50, 105, 196, &c. 1, 6, 20, 50, 105, 196, &c., the given series. 5, 14, 30, 55, 91, &c., 1st differences. 2. Required the several order of differences of the series of 12, 22, 32, 42, 52, &c. 1, 4, 9, 16, 25, &c., the given series. 3, 5, 7, 9, &c., 1st differences. 3. Required the several order of differences of the series of cubes, 13, 23, 33, 4, 53. Ans. 4. Find the order of differences in the series, 4, 1, 16, 32, &c. Ans. PROBLEM II. 338. To find the first term of any order of differences, Let d', d'', d'", d'''', &c., represent the first terms of the 1st, 2d, 3d, 4th, &c., order of differences; then d=—a+b, d'=a -2b+c, d'"-a+3b-3c+d, d'"-a-4b+6c-4d+e, &c.; from which it is obvious that the coefficients of the several terms of any order of differences are respectively the same as those of the terms of an expanded binomial, and are obtained in the same manner; for the terms that are subtracted are actually added, but with contrary signs. Hence we infer that d", or the first n- -1 difference of the nth order of differences, is anb±n. 2 n−1 n2¿±, &c., to n+1 terms; in which formula the cFn. 2 n- -2 3 upper signs must be taken when n is an even number, and the under when n is an odd number. 5. Required the first of the fifth order of differences of the series 6, 9, 17, 35, 63, 99, 148, &c. Let a, b, c, d, e, f, &c. 6, 9, 17, 35, 63, 99, 148, &c., and n(n-1) + = n(n−1)(n−2),__n(n−1)(n—2)(n—3), 2.3 n(n−1)(n—2)(n—3)(n—4)ƒ——a+5b— 2.3.4.5 e 6. Required the first of the sixth order of differences of the series 3, 6, 11, 17, 24, 36, 50, 72, &c. PROBLEM III. Ans. -14. 339. To find the nth term of the series a, b, c, d, e, f, &c. As we have found in the last problem that d'= -a+b, therefore b=a+d', and, in the same manner, we find c=a+2d′+d", d=a+3d'+3d''+d", e=a+4d' +6d" +4d""'+d"'"', &c.; whence the nth term is 7. Required the 7th term of the series 3, 5, 8, 12, 17, &c. 3, 5, 8, 12, 17, &c., the given series. 8. Required the 9th term of the series 1, 5, 15, 35, 70, &c. Ans. 495. 9. Required the 10th term of the series 1, 3, 6, 10, 15, 21, &c. Ans. 55. PROBLEM IV. 340. To find the sum of n terms of the series a, b, c, d, e, &c. If we add the values of a, b, c, &c., as found in the last problem, we obtain 2a+d'=a+b, 3a+3d+d′′=a+b+c, 4a+ 6d'+4d"'+d""=a+b+c+d, &c. Wherefore it is evident that 341. When the differences become at last = 0, any term, or the sum of any numbers, can be accurately found; but, when the differences do not vanish, the formulæ in this and the preceding problem give only an approximation, which will come nearer the truth as the differences diminish. 10. Required the sum of 8 terms of the series 2, 5, 10, 17, &c. =8, a=2, d'=3, d′′=2, and d′′"=0. Here n= 7 6 2 8.2.3.2=16+84+112=212= the sum of 8 terms. 11. Required the sum of 100 terms of the series 1, 2, 3, 4, 5, &c. Here 1, 2, 3, 4, 5, 6, &c., given series. 1, 1, 1, 1, 1, &c., 1st difference. 0, 0, 0, 0, &c., 2d difference. Here n=100, a=1, and d=1. na+n. d=100+ 100. (100-1): 1=505 2 n- -1 12. Required the sum of 12 terms of the series, 1, 4, 10, 20, 35. Ans. 1365. 13. Required the sum of n terms of the series 12, 22, 32, 42, 52, 62, 72, &c. Here 1, 4, 9, 16, 25, 36, 49, &c., given series. 3, 5, 7, 9, 11, 13, &c., 1st difference. |