the is performed, of such a value that the assumprocess ed binomial, &c. shall become =0, and transposing some of the first terms, a series will arise, the sum of which will be known as before. Each of which methods, modified so as to render it more commodious in practice, together with several other artifices for the same purpose, will be found sufficiently elucidated in the miscellaneous questions succeeding the following problems. PROBLEM I. Any series being given to find its several orders of dif ferences. RULE. 1. Take the first term from the second, the second from the third, the third from the fourth, &c. and the remainders will form a new series, called the first order of differences. 2. Take the first term of this last series from the second, the second from the third. the third from the fourth, &c. and the remainders will form another new series, called the second order of differences. 3. Proceed, in the same manner, for the third, fourth, fifth, &c. order of differences; and so on till they terminate, or are carried as far as may be thought necessary.* EXAMPLES. 1. Required the several orders of differences of the se ries 1, 2, 32, 42, 52, 62, &c. 1, 4, 9, 16, 25, 36, &c. 3. 5, 7, 9, 11, &c. 1st diff. 2, 2, 2, 2, &c. 2d diff. 0, 0, 0, &c. 3d diff. 2. Required the different orders of differences of the se ries 1, 23, 33, 43, 53, 63, &c. When the several terms of the series continually increase, the differences will be all positive; but when they decrease, the differences will be negative and positive alternately. 3. Required the several orders of differences of the series 1, 3, 6, 10, 15, 21, &c. Ans. 1st, 2, 3, 4, 5, &c. ; 2d, 1, 1, 1, &c. 4. Required the several orders of differences of the series 1, 6, 20. 50, 105, 196, &c. ་ Ans 1st, 5, 14, 30, 55, 91, &c.; 2d, 9, 16, 25, 36, &c.; 3d, 7, 9, 11, &c. ; 4th, 2, 2, &c. 5. Required the several orders of differences of the 1 1 1 1 1 series &c. '2' 4'3' 16' 32' Any series a, b, c, d, e &c. being given, to find the first term of the nth order of differences. RULE. Let stand for the first term of the nth differences. Then will a nb + n. n -1 -e, &c. to n+1 terms, when n is an even 3 3 4. number. n--3 3 e &c. to n+1 terms, when n is an odd number.* *When the terms of the several orders of differences happen to be very great, it will be more convenient to take the logarithms of the quantities concerned whose differences will be smaller; and when the operation is finished; the quantity answering to the last logarithm may be easily found. EXAMPLES. 1. Required the first term of the third order of differences of the series, 1, 5, 15, 35, 70, &c. Here a, b, c, d, e, &c. =1, 5, 15, 35, 70, &c. and n=3. 2. Required the first term of the fourth order of differences of the series 1, 8, 27, 64, 125, &c. Here a, b, c, d, e, &c. =1, 8, 27, 64, 125, &c. and ʼn≈4. n-1 n 1n- -2 n 1 Whence a-nb+n. C ・d+n. 2 2 3 2 n-2n- -3 3 4 e-a-4b+6c-4d+e=1-32+162-256+125 =0; so that the first term of the fourth order is 0. 3. Required the first term of the eighth order of differences of the series, 1, 3, 9, 29, 81, &c.* Ans. 256. 4. Required the first term of the fifth order of differ 1 1 1 1 1 1 ences of the series, 1, 2' 4' 8' 16' 32' 64' PROBLEM III. To find the nth term of the series, a, b, c, d, e, &c. when the differences of any order become at last equal to each other. , RULE. Let d', d', d'", div, &c. be the first of the several orders of differences, found as in the last problem. The labour in questions of this kind may be often abridged, by putting ciphers for some of the terms at the beginning of the series; by which means several of the differences will be equal to 0, and the answer on that account, obtained in fewer terms. EXAMPLES. 1. It is required to find the twelfth term of the series 2, 6, 12, 20, 30, &c. 2, 6, 12, 20, 30, &c. 4, 6, 8, 10, &c. Here 4 and 2 are the first terms of their differences. Let, therefore, 4=d', 2=d', and n=12. 44+110=156 15th term, or the answer required. 2. Required the twentieth term of the series, 1, 3, 6, 10, 15, 21, &c. 1, 3, 6, 10, &c. Here 2 and 1 are the first terms of the differences. Let, therefore, 2=d, 1=d", and n=20. 38+171=210=20th term required. 3. Required the fifteenth term of the series, 1, 4, 9, 16, 25, 36, &c. Ans. 225. 4. Required the twentieth term of the series, 1, 8, 27, 64,125, &c. Ans. 8000. 5. Required the thirtieth term of the series, 1, 1 1 To find the sum of n terms of the series, a, b, c, d, e, *When the differences in this or the former rule are finally =0, any term, or the sum of any number of the terms, may be accurately determined; but if the differences do not vanish, the result is only an approximation; which, however, may be often very usefully applied in resolving various questions that may occur in this branch of the subject, and which will become continually nearer the truth as the differences diminish. &c. when the differences of any order become at last equal to each other. RULE. Let d', d", d", div, &c. be the first of the several or 1. Required the sum of n terms of the series, 1, 2, 3, 4, 5, 6, &c. Here 1, 2, 3, 4, 5, 6, &c. 1, 1, 1, 1, 1, &c. 0, 0, 0, 0, &c. Where 1 and 0 are the first terms of the differences, Let, therefore, a=1, d'1, and d'=0. Then will na+n.”—'d=n+ terms, as required. n2. -n n2 + n = sum of n 2 2. Required the sum of n terms of the series, 1o, 23, 32, 42, 52, &c., or 1, 4, 9, 16, 25, &c. Here 1, 4, 9, 16, 25, &c. 3, 5, 7, 9, &c. Where 3 and 2 are the first terms of the differences. Let therefore, a=1, d=3, and d'=2. |