In this result we observe that the exponents of both a and x are increased by 1 in each term, and there is still one term without and another without a. Before the terms of the product were added, there were twice as many terms in the product as in the multiplicand, but they have all united two by two except the first and last. The terms Can-3x and Fax- have not united with any others, but it is evident that they would have done so, if all the terms could have been written. There is then one more term in this power than in the last.

The coefficient of the first term is still 1. That of the second is the sum of the coefficients of the first and second terms of the multiplicand, viz. 1+A. That of the third is the sum of the coefficients of the second and third terms of the multiplicand, viz. A + B ; &c.

The above formula shows that if the law above mentioned is true for one power, it will be so for the next higher power. We have seen that it is true for the 5th power, therefore it will be true for the 6th; being true for the 6th, it will be so for the 7th, &c.

Let the coefficients of several of the first powers be written without the letters, forming them by the above principle.

First observe that (a + x)° 1.

=

Adding 0 to this 1 gives 1, and then O again on the other side gives 1. Hence we have 1, 1, for the coefficients of the first power.

Adding 0 to the first 1 gives 1; adding 1 and I gives 2, and then 1 and 0 are 1. Hence the coefficients of the second power are 1, 2, 1.

Again, 0+1=1; 1 + 2 = 3; 2+1 = 3; and 1 + 0 1. Hence 1, 3, 3, 1, are the coefficients of the third power. Again, 0+1=1; 1+3=4; 3+3=6; 3+1=4; and 101. Hence 1, 4, 6, 4, 1, are the coefficients of the fourth power.

=

Again, 0+1=1; 1+4= 5; 4+6= 10; 6+4=10; 4+1 = 5; and 1+0=1. Hence 1, 5, 10, 10, 5, 1, are the coefficients of the 5th power, &c.

1

8 28 56 70 56 28 8 I

1 9 36 84 126 126 84 36 9 1

1 10 45 120 210 252 210 120 45 10 1

Here we observe that the first row of figures taken obliquely downward is the series of numbers 1, 1, 1, &c.

The second row is the series of natural numbers, 1, 2, 3, 4, 5, &c. whose differences are 1.

The third row is the series 1, 3, 6, 10, 15, &c. whose differences are the last series, viz. 1, 2, 3, 4, &c.

The fourth row is the series 1, 4, 10, 20, 35, &c. whose differences are the last series, viz. 1, 3, 6, 10, &c. Each successive row is a series, whose differences form the preceding row.

We may observe farther, that the coefficient of the second term of any power is the term of the series 1, 2, 3, 4, &c. denoted by the exponent of the power. That of the second power is the second term; that of the third power, the third term; that of the nth power, the nth term. But this being the series of natural numbers, the number which denotes the place of the term is equal to the term itself, so that the coefficient of the second term will always be equal to the exponent of the power.

The coefficient of the third term of any power is the term of the series 1, 3, 6, 10, &c. denoted by the exponent of the power diminished by 1. That of the third power is the second term, that of the fourth power the third term, that of the nth power the (n-1)th term, &c.

The coefficient of the fourth term of any power is the term of the series 1, 4, 10, 20, &c. denoted by the exponent of the power diminished by 2. That of the fourth power is the second term, that of the fifth power is the third term,

that of the nth power is the (n-2)th term. And so on as we proceed to the right, the place of the term in the series is diminished by 1.

We may observe another remarkable fact, the reason of which will be manifest on recurring to the formation of these series. We shall take the 7th power for an example, though it is equally true of any other.

The coefficient of the second term, viz. 7, is the sum of 7 terms of the preceding series 1, 1, 1, &c. and was in fact formed by adding them.

The coefficient of the third term, 21, is the sum of the first six terms of the preceding series, 1, 2, 3, &c. and was actually formed by adding them, as may be seen by referring to

the formation.

The coefficient of the fourth term, 35, is the sum of the first five terms of the preceding series, 1, 3, 6, 10, &c. and was formed by adding them.

The same law continues through the whole. If now we can discover a simple method of finding the sums of these series without actually forming the series themselves, it will be easy to find the coefficients of any power without forming the preceding powers. This will be our next inquiry.

Summation of Series by Differences.

A series by differences is several numbers arranged together, the successive terms of which differ from each other by some regular law.

I call a series of the first order that, in which all the terms are alike, as 1, 1, 1, 1, &c. 3, 3, 3, 3, &c. a, a, a, a, &c. In these the difference is zero.

The sum of all the terms of such a series is evidently found by multiplying one of the terms by the number of terms in the series. Every case of multiplication is an example of finding the sum of such a series.

The sums of a number n of terms of any series a, a, a, &c. may be expressed

A series in which the terms increase or diminish by a constant difference, I call a series of the second order. As 1, 2, 3, 4, 5, &c. 3, 6, 9, 12, &c. or 12, 9, 6, 3. A series of

this kind is formed from a series of the first order.

The

differences between the successive terms form the series from which it is derived.

At present I shall examine only the series of natural numbers 1, 2, 3, 4,

...n.

This series is formed as follows:

0+1=1

1+1=2

1+1+1= 3

1 +1 +1 +1 = 4

1+1+1+1+1=5&c.

The sum of any number n of terms of the series 1, 1, 1, 1, &c. is equal to the nth term of the series 1, 2, 3, 4, &c. Write down two of these series as follows and add the corresponding terms of the two together.

1, 2, 3, 4, 5

5, 4, 3, 2, 1

6, 6, 6, 6, 6

3,

1

1, 2, 4,... (n-3), (n-2), (n−1,) n n, (n-1), (n-2), (n-3)....4, 3, 2, (n+1), (n+1), (n+1), (n+1). . .(n+1), (n+1), (n+1), (n+1) The 6th term of the series is 6, and it appears that 5 times 6 will be twice the sum of 5 terms of the series.

The (n+1)th term of the series 1, 2, 3, 4, &c. is n + 1. It appears that n times (n + 1) will be twice the sum of n terms of the series.

The sum of any number n of terms may be expressed

The rule expressed in words is; To find the sum of any number of terms of the series 1, 2, 3, 4, &c. find the next succeeding term in the series, and multiply it by the number of terms in the series, and divide the product by 2.

1+2+3+4+5+6=15+621 &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.

6 X

1+2+3+4+ 5+ 6 =

6 X 7
2

=21 6th 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, are 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 5 terms, or 1+ 2+ 3+ 4+ 5 = 5 x 6.

2