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XLIII. The principle of summing these series may be proved generally as follows:

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Let 1, a, b, c, d .... I be a series of any order, such that the sum of n terms may be found by multiplying the (n + 1)th term by n, and dividing the product by m. If I is the (n + 1) th term, and s the sum of all the terms, we shall have by hy pothesis

nl

and ms nl.

m

That is, n l will be m times the sum of the series. The next higher series will be formed from this as follows:

1

= 1st term.

1 + a
1+ a +6
1 tatb toc
1+ a + b tictd

= 2d
= 3d
= 4th 5
= 5th

= nth

1+a+b+c+d+....k
1 + a + b + c + d +....k+1= (n + 1)th.

The first term 1 of the original series 1, a, b, &c., forms the first, term of the new series; the sum of the first two forms the second term; the sum of the first three forms the third term, &c., and the sum of (n + 1) terms forms the (n + 1)th term.

Let'the series forming the (n + 1)th term, be written » times, one under the other, term for term. And let a line be drawn diagonally, so that the first term of the first row, the first two of the second row, and in terms of the nth row may be at the left, and below the line.

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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

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The sum of four terms, or 1 + a +6+

4 d

The sum of n terms, or 1+a+b+c+d+

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m. 1

= 26

- Multiplying both members of the above equations by m:

=la
m (1 + a)
m (1 ta+b)

= 32
m (1 + a + b + c)

=4d mm (1 to a +b+c+d+...k)

=nl 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 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.

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n

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When the series is of the first order, the sum of n terms is n.1 cor 1 1

n +1 The sum of (n + 1) terms of this series is

This is

1 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)

Х 2 Х 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 Х 4

Hence for the series of the order r we have this formula:

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

1 X 2 X 3 X 4 X.... 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.

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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 ;

a, x, ao x*, a* a*, ao x*, amo, a , 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.

6 x 7

= 21.

2 The coefficient is 21.

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 preced ing 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.

5 X 21

3

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= 35.

The cocfficient is 35.

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 coeificient last found, viz. 35.

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