Εικόνες σελίδας
PDF

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. w

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 equal'y 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.

XLII. Summation of Series by Differences.

It is not my purpose at present to enter very minutely into the theory of series. I shall examine only a few of the most simple of them, and those principally with a view of demonstrating the binomial theorem.

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 feries of the first order hat, 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 sum s of a number n of terms of any series a, a, a, &c. is expressed

[ocr errors][ocr errors][ocr errors][ocr errors]

A series in which the terms increase or diminish by a constant difference, is called 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, . . . . . . . 7t.

[merged small][ocr errors]

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 •

18 + 6, 6, 6, 6, 6

[ocr errors][ocr errors]

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 s' of any number n of terms may be expressed thus.

1. 2 It is frequently convenient to use the same letter in similar situations to express different values. In order to distinguish it in different places, it may be marked thus, s, s', s”, s”, which may be read s, s prime, s second, s third, &c. How many times does the hammer of a clock strike in 12 hours ?

In this example n = 12 n + 1 = 13. 12 × 13 = 78. 1 X 2 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.

[ocr errors]

The same thing may be proved in another form which is more conformable to the method that will be used for the series of the higher orders.

Suppose it is required to find the sum of the first five terms of the series.

The sixth term of \he series is the sum of 6 terms of the series, 1, 1, 1, &c. thus

[ocr errors]

Let this series be written down five times, one under the other, thus.

[subsumed][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors][ocr errors]

If this series be divided by a line passing diagonally through it, so that the part below and at the left of the line may contain one term of the first series, two of the second, three of the third, four of the fourth, and five of the fifth ; the terms so separated will form the first five terms of the series 1, 2, 3, &c. There will be the same number of terms above and at the right of the line, which will form the same series, if the terms be added vertically instead of horizontally.

1, 1, 1, 1, 1, - 1

It is easy to see that this series continued to any number of terms will be formed twice over in this way, if the number of series written under each other is equal to the number of terms required and the number of terms in each series exceed the number of terms by one. And the reason of it is manifest from the manner in which the two series are formed.

Hence n times the series consisting of n + 1 terms of the series 1, 1, 1, 1, &c. will be twice the sum so of n terms of the Series 1, 2, 3, 4, &c.

[ocr errors]
[graphic]

A series of the third order is one, the difference of the successive terms of which is a series of the second order. I shall consider only the series formed from the series 1, 2, 3 &c.

[ocr errors][ocr errors]

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.

[ocr errors]

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.

3. 2, 3, 4, 5, 2,

1 1,

6 6 6 6 6

1, 2, 3, 1, 2, 3, 1, 2, 3,

[graphic]
« ΠροηγούμενηΣυνέχεια »