By the rule given above for finding the sum of the series 1, 3, 6, 10, &c. The sum of one term, or 1 1X3 -1. 3 The sum of four terms, or 1+3+6 +10= 4 X 15 = 20. 3 The sum of five terms, or 1+3+6+10+15=5 X 21 = 35. colx The five 21s are 3 times 1+3+6 +10 + 15. The four 15s are 3 times 1+3+ 6 + 10 and so of the rest. It is easy to see that this principle will extend to any number of terms. Therefore to find the sum of n terms of the series 1, 4, 10, 20, &c., multiply the (n + 1)th term of the series by n, and divide the product by 4, and the quotient will be the sum required. But the (n + 1)th term of this series is equal to the sum of (n + 1) terms of the preceding series. The nth term of the preceding series being n(n + 1)(n+2) 1 X 2 X 3 the (n + 1)th term will be (n + 1)(n+2) (n + 3). 1 X 2 X 3 gl! = n(n+ 1) (n + 2) (n + 3). Х 4 DIVOR CALIFORNIA XLIII. 217 XLIII. The principle of summing these series may be proved generally as follows: Let 1, a, b, c, d ......l be a series of any order, such that the sum of n terms may be found by multiplying the (n + 1)th term by n, and dividing the product by m. If I is the (n + 1) th term, and s the sum of all the terms, we shall have by hypothesis That is, n l will be m times the sum of the series. The next higher series will be formed from this as follows: = nth 1 + a + b +c+ d +....k The first term 1 of the original series 1, a, b, &c., forms the first term of the new series; the sum of the first two forms the second term; the sum of the first three forms the third term, &c., and the sum of (n + 1) terms forms the (n + 1)th term. Let the series forming the (n + 1)th term, be written n times, one under the other, term for term. And let a line be drawn diagonally, so that the first term of the first row, the first two of the second row, and n terms of the nth row may be at the left, and below the line. 19 The terms below and at the left of the line, form n terms of the new series. It is now to be shown that the terms above, and at the right of the line, are equal to m times those below, and, consequently, that the whole together are equal to m + 1 times n terms of the new series. By the hypothesis nl The sum of n terms, or 1+a+b+c+d+..k= m m. 1 Multiplying both members of the above equations by m: =la m (1 + a) = 26 =30 =nl Hence it appears, that a is m times 1 ; 2 b is m times (1 + a) &c.; and nl is m times (1 +a+b+c+d+.... k); that is, the part above and at the right of the line, is m times the part at the left and below; consequently the whole, or n tim :s the (n + 1)th term of the new series, will be (m + 1) ti jes the sum of n terms of the same series. We have already examined all the series as far as the fourth order, and have found the above hypothesis true so far. Let us suppose the series 1, a, b, &c. to be a series of the fourth order, in which we have found that the sum of n terms may be obtained by multiplying the (n + 1)th term by'n, and dividing the product by 4; in this case m is equal to 4. The series formed from this will be a series of the 5th order, and m + 1 = 4+1=5. Therefore by the above demonstration it appears that the sum of n terms of a series of the 5th order may be obtained by multiplying the (n + 1)th term by n, and dividing the product by 5. If now the series, 1, a, b, &c., be considered a series of the 5th order, m = 5 and m +1=6. Hence the same principle extends to the 6th order. If then we continue to make 1, a, b, &c., represent one series after another in this way, we shall see that the principle will extend to any order whatever of this kind of series. We have then this general rule; To find the sum of n terms of a series of the order denoted by r, derived from the series 1, 1, 1, &c., multiply the (n + 1)th term of the series by n and divide the product by r. Also, the nth term of the series of the order r, is equal to the sum of n terms of the series of the order r 1. n or When the series is of the first order, the sum of n terms is n.1 1 1 The sum of (n + 1) terms of this series is " + 1 , This is 1 the (n + 1)th term of the series of the second order. This multiplied by n and divided by 2 gives the sum of n terms of the series of the second order: n (n + 1) 1 X 2 The sum of (n + 1) terms of the same series is (n + 1)(n + 2) 1 X 2 This is the (n + 1)th term of the series of the third order. This multiplied by n and divided by 3 gives the sum of n terms of this series : n(n + 1) (n + 2) 1 X 2 X 3 The sum of (n + 1) terms of the last series is (n + 1)(n+ 2) (n + 3) 1 x 2 Х 3 This is the (n + 1)th term of the series of the fourth order. This multiplied by n and divided by 4 gives the sum of n terms of the series of the fourth order : n(n + 1) (n + 2)(n+3) Hence for the series of the order r we have this formula : n (n + 1) (n + 2)(n + 3)....(nto -1). T We have examined only the series formed from the series 1, 1,1, 1, &c., which are sufficient for our present purpose. The principle may be generalized so as to find the sum of any series |