A series of the third order is one, the difference of the succes- . sive terins of which is a series of the second order. I shall consider only the series formed from the series 1, 2, 3 &c. 1+2+3+4+5+6=15+6=21, &c. The first term of the series 1, 2, 3, &c. forms the first term the sum of the first two terms forms the second; the sum of the first three forms the third term, &c. and the sum of n terms will form the nth term of the series 1, 3, 6, 10, &c. Let it be required to find the sum of the first five terms of the series 1, 3, 6, 10, 15, 21, &c. The sixth term of this series is the sum of the first 6 terms of the series 1, 2, 3, &c. 1+2+3+4+5+6= 6 x 7 =216th term. Write this series five times one under the other, and draw a line diagonally so as to leave on the left and below, the first term of the first, the first two of the second, the first three of the third, &c. and the first five of the fifth. The figures so cut off form the first five terms of the series 1, 3, 6, 10, 15, &c. the sum of which we wish to find. It will now be shown that the sum of the terms on the right and above the line, is equal to twice the sum of those below and at the left. By the rule given above for finding the sum of the series 1, 2, 3, &c. The sum of 1 term, or 1 The sum of 2 terms, or 1+2 The sum of 3 terms, or 1+2+3 The sum of 4 terms, or 1+2+3+4 The sum of 5 terms, or 1+2+3+4+5 Hence 2 (1) =1X2 Since the part below the line forms the series whose sum is required, and the part above the line is equal to twice that below, both parts together are equal to three times the series 1, 3, 6, 10, 15. Therefore if 21, which is the next term in the series, and which is also the sum of the series 1, 2, 3, 4, 5, 6, be multiplied by 5, the number of terms to be summed, and divided by 3, the quotient will be the sum of the series required. It is easy to see that if the series 1, 2, 3,... (n + 1) be written ʼn times and divided by a line like the above, the part below the line will form n terms of the series 1, 3, 6, 10, &c. And the part above the line will be equal to twice the part below, because the sum of n terms of the series 1, 2, 3, &c. is n (n + 1) Therefore to find the sum of n terms of the series 1, 3, 6, 10, multiply the (n+1)th term of that series by n and divide by 3, and the quotient will be the sum required. But the (n+1)th term of the series is equal to the sum of (n+1) terms of the series 1, 2, 3, 4, &c. The nth term of this series being " (n+1), the (n+1)th term will.be n 1 x 2 (n + 1) (n + 2) 1 X 2 This being multiplied by n, the number of terms, and divided by 3, gives n (n + 1) (n + 2) 1 X 2 X 3 Hence the sum s" of n terms of the series will be expressed thus, A series of the fourth order is one, the difference of whose terms is a series of the third order. I shall at present consider only the one formed from the series 1, 3, 6, 10, 15, &c. 1+3+6+10+15+21=35+21=56 The first term of the series 1, 3, 6, &c. is the first term of the new series; the sum of the first two terms forms the second; &c. the sum of n terms will form the nth term of the new series. It is required to find the sum of five terms of this series. The sixth term of this series is equal to the sum of the first six terms of the preceding. 1+3+6+10+15+21= 6 X 7 X 8 1 X 2 X 3 = 56. Write this series five times, one under the other, and separate it into two parts by a line drawn diagonally in the same manner as was done with the last series. The terms below the line will form the series whose sum is required, and the terms above the line will be equal to three times those below. That is, the whole will be four times the sum required. By the rule given above for finding the sum of the 'series 1, 3, 6, 10, &c. The sum of four terms, or 1 + 3 + 6 + 10 = = 10. 8x10= - 3 16 = The sum of five terms, or 1+3+6+10+15= 5 × The five 21s are 3 times The four 15s are 3 times 3 5 × 21 = 20. = 35. 1+3+6+10 + 15. 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 This being multiplied by n and divided by 4, gives |