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 n lis 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 ʼn 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 ʼn 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 n terms is

n

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)

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 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 n 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 :

n (n + 1) (n + 2) (n + 3) . . . . (n + r−1) ̧

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 ti now be required to find the 7th power of a +x. The letters without the coefficients stand thus;

a2, a® x, a3 x2, aa x3, a3 x1, a3 x3, α xo, x7.

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. 19*

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 3d term of this series is the coefficient last

the 6th order.

found, viz. 21.

2 X 21 = 7.

6

The coefficient is 7.

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 a2 x3 +7 axˆ+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

the term to be multiplied, because the number of terms always diminishes by 1 for the successive coefficients, and the place of the term always marks the order of the series of which the sum is to be found.

Hence is obtained the following general rule.

Knowing the coefficient of any term in the power, the coefficient of the succeeding term is found by multiplying the coefficient of the known term by the exponent of the leading quantity in that term, and dividing the product by the number which marks the place of that term from the first.

The coefficient of the first term, being always 1, is always known. Therefore, beginning with this, all the others may be found by the rule.

It may be farther observed, that the coefficients of the last half of the terms, are the same as those of the first half in an inverted order. This is evident by looking at the coefficients, page 275, and observing that the series are the same, whether taken obliquely to the left or to the right.

It is also evident from this, that a +x is the same as x+", and that, taken from right to left, a is the leading quantity in the same manner as a is the leading quantity from left to right.

Hence it is sufficient to find coefficients of one half of the terms when the number of terms is even, and of one more than half when the number is odd. The same coefficients may then be written before the corresponding terms counted from the right.

In the above example of the 6th power, the coefficients of the first four terms being found, we may begin on the right and put 6 before the second, and 15 before the third, and then the power is complete.

Examples.

1. What is the 7th power of a + x?

Ans. a1 +7 a x +21 a3 x2 + 35 a* x3 +35 a3 xa + 21 a2 x3

+7ax® + x2.