of its terms by 2, when put under the form 2 = m — (m − 2) for the multiplier of the first term (m-1) of the series, and under the form 2 = (m − 1) -1)-(m3) for the multiplier of the second term (m2), and 2 = (m − 2) — (m-4) is the multiplier of the third term (m 3), and so on; so that we continually change the form of the multiplier without changing its value. - The product of the first term of the series and its multiplier is m(m − 1) — (m 1) — (m − 1)(m2); and if in this we change m into m-1, we shall have (m-1) (m − 2) — (m − 2) (m—3) for the product of the second term of the series and its multiplier; and if in the last product we change m into m - 1, we get (m − 2)(m — 3) — (m − 3) (m — 4), and so on to the last product, which is expressed by [m (m2)]. [m(m − 1)] — [m — ( m− 1)] . [m — m]; whose last term is 0, since its factor mm equals 0. Hence, adding the preceding products, by putting corresponding terms in vertical columns under each other, we get the following results. m(m—1)—(m—1)(m—2) (m—1)(m—2)—(m—2)(m—3) (m—2)(m—3)—(m—3)(m—4) (m—3)(m—4)— and so on to the last product inclusive, whose sum is m(m 1); since the terms in vertical columns destroy each other, and that the last term of the last product is 0. Hence, since twice the series is transformed to m(m — 1), we shall have (m − 1) + (m − 2) + (m − 3) + (m − 4) + m(m − 1) +1= 2 " (2), as required. Because m= (m −1) + 1 = the sum of the extreme terms of the series, and that m− 1 = the number of terms, we deduce from (2) the following rule for finding the sum of the terms of the series. RULE. The sum of the terms of the series equals half the product of the sum of the extreme terms by the number of terms. + Ex. 2.-To transform the series (m − 1) (m − 2) + (m − 2) (m − 3) + (m − 3) (m − 4) + (m − 4) (m −5) + [m — (m − 2)]. [m - (m-1)], (1), in such a way that its sum can be found. Here we multiply each term of the series by 3, which is to be put under the form m — (m − 3) for the multiplier of the first term (m-1) (m2) of the series; and (m-1) — (m 4) is the multiplier of the second term (m2)(m − 3); 5) is the multiplier of the third term 3) (m — 4), and so on to [m- (m— 3)] — (m — m), which is the multiplier of the last term. - and (m (m 2) - (m The product of the first term of the series and its multiplier is m(m − 1)(m − 2) — (m − 1)(m − 2) (m · 1)(m2) (m3), and putting in this m-1 for m, we have (m1)(m2) (m-2)(m 3) — ( m -2) 4) for the product of the second term of the series and its multiplier; and changing in this last produet m into m 1, we have (m2) (m − 3) (m — 4) - (m 3) (m-4) (m -4) (m5) for the product of the third term of the series and its multiplier, and so on, to the product of the last term of the series and its multiplier, which is [m — (m −3)]. [m - (m2)]. [m- (m-1)], by omitting mm = 0 in the multiplier. -- Hence, if we add all the preceding products, as in the preceding example, their sum will be reduced to m(m − 1)(m − 2), which is of course equal to 3 times the sum of all the terms of the given series; consequently we have (m −1)(m − 2) + (m − 2) (m − 3) + (m − 3) (m − 4) + (m — 4) (m m(m—1)(m—2) Ex. 3. To transform the series (m − 1)(m − 2) (m − 3) + (m − 2) (m − 3) (m − 4) + (m — 3) (m — 4) (m − 5) + +[m― (m3)]. · (m − 3)] . [m — (m − 2)] . [m — (m − 1)] (1). Here we must multiply each term by 4, which is greater by a unit than the number of factors in each of the terms of the given series. For the multiplier of the first term of the series we must put 4 m (m-4), and for the multiplier of the second = term of the series we must put 4 (m1) (m5), and 4 = (m − 2) — (m — 6) is the multiplier of the third term, = — = and so on. Hence, (as in the preceding examples), we easily get (m-1)(m − 2) (m − (m −3) (m − 4) (m − 5) 3) + (m − 2) (m − 3) (m + . . . . + [m — (m −.3)] × [m .... (m − 2]. [m — (m − 1)] m(m − 1)(m − 2) (m — 3), as required. = 4 3), (2), Universally, if n is any positive integer, and we have the series of successive products, each containing n factors, as (m-1) (m2) (m3) × .... × (m − n) + (m — 2) (m — 3) (m −4) × . . . . × [m − (n + 1)] + (m − 3) (m — 4) (m —— 5) × .... × [m − (n + 2)] + (m — 4)(m −15) (m − 6) × — .... × [m− (n + 3)]+ + [m ~ (m ~ n)] . [m — (m—n+1)]. [m― (m − n + 2)] × .... × [m (m-1)], such that any term of the series is deduced from the preceding term by changing m into m1; then the multiplier of each term must be n + 1, which is greater than the number of factors in each term of the series by a unit. — For the multiplier of the first term of the series, we must put n + 1 = m (mn - 1); and changing m into m−1, we have n + 1 = (m − 1) — (m-n-2) for the multiplier of the second term of the series; and changing in this multiplier m into m-1, we have (m- 2) — (m n-3) for the multiplier of the third term of the series, and so on. Consequently, if we denote the sum of all the terms of the series (for brevity) by S, we shall have (as before) S = Ex. 4.-Using m to represent any positive integer, as heretofore, it is required to find the sum of the series m2 + (m − 1)2 + (m − 2)2 + (m − 3)2 + . . . . + 13 (1). Since m2 = m + m2 - m, we get m2 = m + m(m − 1); and changing m into m-1, we have (m- 1)2 = (m — 1) + (m − 1)(m − 2); and changing m into m— 1, we get (m − 2)2 = (m − 2) + (m − 2) (m3), and so on. Consequently, by adding the preceding results, we get m2 + (m-1)+(m-2)2 + (m − 3)2 + + 12 = m + (m — 1) + (m − 2) + (m − 3) + +1+m(m − 1) + (m − 1) + 2 x 1. (2) of Ex. 1 we put m + 1, it will give m + (m - 1) + (m − 2) + (m − 3) + . . . . + 1 = m(m + 1); and if in 2 (2) of Ex. 2, we put m +1 for m, we shall get m(m—1) +(m-1)(m2) + (m − 2) (m − 3) + — + 2x1 = (m1)m(m + 1) 3 Hence, we deduce m2 + (m1)2 + (m − 2)2 + (m — 3)a m(m+1) (m−1)m(m+1) __ m(m+1)(2m+1) + .... +12= (2), as required. 2 = Ex. 5. To find the sum of the series m3 + (m − 1)3 + (m − 2)3 + (m − 3)3 + . . . . + 13 (1). Since m3 m + m3 — m = m + m(m2 - 1), and that m2 -1= (m + 1)(m-1), we of course get m3 = m + (m + 1) m(m − 1); and changing m into m−1, we have (m — 1)3 = (m − 1) + m(m − 1)(m − 2); and changing in this, m into m-1, we get (m − 2)3 = (m − 2) + (m − 1)(m − 2) (m3), and so on. Adding the preceding results, we have m3 + (m − 1)3 + (m − 2)3 + (m − 3) + . . . . +13 = m + = (m · 1) + (m − 2) + (m −3) + ...+1+(m + 1)m(m − 1) +m(m − 1)(m 1) (m − 2) + (m − 1) (m − 2) (m − 3) + . . . . + 3 × 2 × 1. As in the last example, we have m + (m − 1) + (m - -2) for m in (2) of Ex. 3, we get (m + 1). m(m − 1) + m(m − 1) 2) + (m − 1) (m − 2) (m − 3) + .... + 3 × 2 × 1 = (m + 2) (m + 1)m(m — 1) to signify the product of 1 and 2; also 2.3 is used to express the product of 2 and 3, and so on; also n is used for a whole number, which contains as many units as there are units in the number of any term, when the terms are counted from the left to the right. To find the sum of the series, we multiply each term by 1; when put under the forms 1=2-1 for the multiplier of the first term, 132 for the multiplier of the second term, 143 for the multiplier of the third term, and so on, to (n+1)-n, which is the multiplier of the term whose number is n. and the product of the second term and its multiplier is the product of the third term and its multiplier is and so on, to the term whose number is n, whose product by its multiplier is 1 1 n n+1 Hence, by adding erasing the terms that are the preceding products, and numerically equal and affected with opposite signs, which Here we must multiply each term by 2, which is less by a unit than the number of factors in the denominator of each term. We must put 231 for the multiplier of the first term; and 24-2 for the multiplier of the second term; and 25-3 for the multiplier of the third term, and so on, to (n+2)-n, which is the multiplier of the term whose number is n. The product of the first term and its multiplier is 1 1.2 is the product of the second term |