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

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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

1 1,1, 1, 1, 1, . 1, 1, 1, 1, 1, 1

1, 1, 1, 1, 1, 1

1, 1, 1, 1, 1,11 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 s of n terms of the series 1, 2, 3, 4, &c.

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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.

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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.

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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.

1, 2, 3, 4, 5, 6
1, 2, 3, 4, 5, 6
1, 2, 3, 4, 5, 6
1, 2, 3, 4, 5, 6
1, 2, 3, 4, 5, 6

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, is 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 1 term, or 1

1 X 2

The sum of 2 terms, or 1+2

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Hence 2 (1)

2 (1+2)
2 (1+2+3) .
2 (1+2+3+4) = 4 X

2 (1+2+3+4+5) = 5 x 6
That is, the 2 is twice the 1, '.
The two threes are twice (1 +2),
The three fours are twice (1 +2+3),
The four fives are twice (1 + 2 + 3+ 4), and
The five sixes are twice (1 +2 +3+4+5).

Since the part below the line forms the series whose sum is required, and the part above the line is equal to twice that below, both parts together are equal to three times the series 1, 3, 6, 10, 15. Therefore if 21, which is the next term in the series, and which is also the sum of the series 1, 2, 3, 4, 5, 6, be multiplied by 5, the number of terms to be summed, and

Jivided 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)

1x 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 + 1), the (n + 1)th term will be

si x2

(n + 1)(n +2)

1 X 2 This being multiplied by n, the number of terms, and divided by 3, gives

n(n+1)(n + 2)

1x2 X 3 Hence the sum gel of n terms of the series will be expressed

thus,

dln(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.

Formation.

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0+1

. . = 0 + 1 = 1 1 +3

= 1 + 3 = 4 . 1+3+ 6

= 4 + 6 = 10 1+3+6 + 10

= 10 + 10 = 20 1+3+ 6 + 10 +- 15 = 20 + 15 = 35

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 =

= 56. .

1 * 2 x 3 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 done 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.

1,\ 3, 6, 10, 15, 21

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