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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.

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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 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) Ix 3 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 #9, the (n + 1)th term will be

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This being multiplied by n, the number of terms, and divided by 3, gives - n (n + 1) (n + 2). 1 × 2 × 3 * Hence the sum s” of n terms of the series will be expressed thus, y = n (n + 1)(n+2). 1 × 2 × 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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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 SerleS.

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.

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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 lone 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.

o 6, 10, 15, 21

10, 15, 21

10, 15, 21 1, 3, 6, 10, N15, 21

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

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

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The four 15s are 3 times 1 + 3 + 6 + 10 and so of the rest.

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

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

Let 1, a, b, c, d . . . . . . l be a series of any order, such that the sum of n terms may be found by multiplying the (n + 1)th ter: by ::, 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 hy

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

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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 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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