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divided by 3, the quotient will be the sum of the series required.

It is easy to see that if the series 1, 2, 3,... (n + 1) be written n times and divided by a line like the above, the part below the line will form n terms of the series 1, 3, 6, 10, &c. And the part above the line will be equal to twice the part below, because the sum of n terms of the series 1, 2, 3, &c. is

n (n + 1)

1 X 2

Therefore to find the sum of n terms of the series 1, 3, 6, 10, multiply the (n+1)th term of that series by n and divide by 3, and the quotient will be the sum required.

But the (n+1)th term of the series is equal to the sum of (n + 1) terms of the series 1, 2, 3, 4, &c. The nth term of this series being n (n + 1), the (n + 1)th term will.be 1 x 2

(n + 1) (n + 2)

1 X 2

This being multiplied by n, the number of terms, and divided by 3, gives

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Hence the sum s" of n terms of the series will be expressed thus,

s' _ n (n + 1) (n + 2)

1 X 2 X 3

A series of the fourth order is one, the difference of whose terms is a series of the third order.

I shall at present consider only the one formed from the series 1, 3, 6, 10, 15, &c.

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1+3+6+10+15+ 21 = 35 +21=56

The first term of the series 1, 3, 6, &c. is the first term of the new series; the sum of the first two terms forms the second; &c. the sum of n terms will form the nth term of the new series.

It is required to find the sum of five terms of this series. The sixth term of this series is equal to the sum of the first six terms of the preceding.

1+3+6 +10 + 15 + 21 =

6 X 7 X 8
1 X 2 X 3

= 56.

Write this series five times, one under the other, and separate it into two parts by a line drawn diagonally in the same manner as was one with the last series. The terms below the line will form the series whose sum is required, and the terms above the line will be equal to three times those below. That is, the whole will be four times the sum required.

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By the rule given above for finding the sum of the series 1 3, 6, 10, &c.

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The sum of four terms, or 1+3+6+10=

The sum of five terms, or 1+3+6+10+15=

The five 21s are 3 times,

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1+3+6+10+15.

The four 15s are 3 times

and so of the rest.

1+3+6+10

It is easy to see that this principle will extend to any number of terms.

Therefore to find the sum of n terms of the series 1, 4, 10, 20, &c., multiply the (n + 1)th term of the series by n, and divide the product by 4, and the quotient will be the sum required.

But the (n+1)th term of this series is equal to the sum of (n+1) terms of the preceding series.

The nth term of the preceding series being

n (n + 1) (n + 2)

1 x 2 X 3

the (n+1)th term will be

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

1 X 2 X 3

This being multiplied by n and divided by 4, gives

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XLIII.

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

n

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

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

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1+a+b+c+d+....k

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66

=nth
1+a+b+c+d+.... k +7= (+1)th.
l

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 *econd 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 n 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 n terms of the nth row may be at the left, and below the line.

19

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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

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

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The sum of ʼn terms, or 1+a+b+c+d+..k=

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