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 The sum of one term, or 1 The sum of two terms, or 1+ a The sum of three terms, or 1+a+b The sum of four terms, or 1+a+b+c = The sum of n terms, or 1+a+b+c+d+..k= 1 a m 26 m 3 c m 4 d m nl Multiplying both members of the above equations by m: 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 times the (n + 1)th term of the new series, will be (m + 1) times 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. When the series is of the first order, the sum of ʼn terms is The sum of (n + 1) terms of this series is 1 This is 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) 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 ʼn 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 X 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 ʼn terms of the series of the fourth order : n (n + 1) (n + 2) (n+3) 1 X 2 X 3 X 4 Hence for the series of the order r we have this formula: 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 of the kind, whatever be the original series, and whatever be the first terms of those formed from it. XLIV. Binomial Theorem. Before reading this article, it is recommended to the learner to review article XLI. Let it now be required to find the 7th power of a +x. The letters without the coefficients stand thus ; a2, a3 x, a3 x2, aa x3, a3 x1, a3 x3, α xo‚x”. The coefficient of the first term we observed Art. XLI, is always 1. That of the second term is 7, the exponent of the power, or the 7th term of the series 1, 2, 3, &c. The coefficient of the third term is the sixth term of the series of the third order 1, 3, 6, 10, &c. which is the sum of six terms of the series 1, 2, 3, &c. This sum is found by multiplying the 7th term of the series by 6 and dividing the product by 2. But the 7th term is 7, the coefficient last found. The coefficient of the fourth term is the 5th term of the series 1, 4, 10, &c., or it is the sum of five terms of the preceding series. The sum of five terms of the series 1, 3, 6, &c., is found by multiplying the 6th term by 5 and dividing the product by 3. The 6th term is the coefficient last found, viz. 21. The coefficient of the fifth term is the fourth term of the series of the fifth order 1, 5, 15, &c., or it is the sum of 4 terms of the preceding series. The sum of 4 terms of the series 1, 4, 10, &c. is found by multiplying the fifth term of the series by 4 and dividing the product by 4. The fifth term is the coefficient last found, viz. 35. The coefficient of the 6th term is the 3d term of the series of the sixth order, which is the sum of 3 terms of the series of the 5th order. The sum of 3 terms of this series is found by multiplying the 4th term by 3 and dividing the product by 5. The 4th term is the coefficient last found, viz. 35 The coefficient of the 7th term is the 2d term of the series of the 7th order, which is the sum of two terms of the series of the 6th order. The 3d term of this series is the coefficient last found, viz. 21. The coefficient of the last term is 1, though it may be found by the rule Hence the 7th power of a +x is a2 + 7 ao x + 21 a3 x2 +35 a1x3 +35 a3 x* +21 a2x2 +7 ax® +x2 Examining the formation of the above coefficients, we observe, that each coefficient was found by multiplying the coefficient of the preceding term by the exponent of the leading quantity a in that term, and dividing the product by the number which marks the place of that term. Thus the coefficient of the third term was found by multiplying 7, the coefficient of the second term, by 6, the exponent of a in the second term, and dividing the product by 2, the number which marks the place of the second term. This will be true for all cases, because that exponent must necessarily show the number of terms of which the sum is to be found; the coefficient will always be |