Of the Method by Differences.

.

209. Let a, b, c, d . . &c., represent the successive terms of a series formed according to any fixed law; then if each term be subtracted from the succeeding one, the several re mainders will form a new series called the first order of dif ferences. If we subtract each term of this series from the succeeding one, we shall form another series called the second order of differences, and so on, as exhibited in the annexed

lf, now,

e-4d6c4b+a, &c., 4th.

we designate the first terms of the first, second. third, &c. orders of differences, by d1, da, da, da, &c., we shall

de-4d6c-4b+a, whence ea + 4d, + 6d2 + 4d + dan

This formula enables us to find the (n+1)th term of a

series when we know the first terms of the successive orders of differences.

210. To find an expression for the sum of n terms of the series a, b, c, &c., let us take the series

0, a, a + b, a + b + c, a+b+c+d, &c.

(2)

(3)

Now, it is obvious that the sum of n terms of the series (3), is equal to the (n+1)th term of the series (2).

is

But the first term of the first order of differences in series (2)

a; the first term of the second order of differences is the same as d in equation (1). The first term of the third order of differences is equal to do, and so on.

Hence, making these changes in formula (1), and denoting the sum of n terms by S, we have,

When all of the terms of any order of differences become equal, the terms of all succeeding orders of differences are 0, and formulas (1) and (4) give exact results. When there are

no orders of differences, whose terms become equal, then formulas do not give exact results, but approxiinations more or less exact according to the number of terms used.

EXAMPLES.

1. Find the sum of n terms of the series 1.2, 2.3, 3.4, 4.5, &c.

Hence, we have, a 2, d, 4, d2 = 2, da, da, &c., equal

0,

0.

to 0.

Substituting these values for a, d1, do, &c., in formula (4),

2. Find the sum of n terms of the series 1.2.3, 2.3.4,

We find a 6, d1 = 18, d2 = 18, d2 = 6, d1 = 0, &c.

=

Substituting in equation (4), and reducing, we find,

3. Find the sum of n terms of the series 1, 1+2, 1+2+3,

4. Find the sum of n terms of the series 12, 22, 32, 42, 52, &c.

We find, a = 1, d1 = 3, d2 = 2, ds = 0, d1 = 0, &c., &c.

Substituting these values in formula (4), and reducing, we find, n(n + 1) (2n + 1) S=

5. Find the sum of n terms of the series,

1. (m+1), 2(m+2), 3 (m+3), 4 (m + 4), &c.

We find, am + 1, d1 = m + 3, d2 = 2, ds = 0, &c.;

The last three formulas deduced, are of practical application in determining the number of balls in different shaped piles.

First, in the Triangular Pile.

211. A triangular pile is formed of succcessive triangular layers, such that the number of shot in each side of the layers, decreases continuously by 1 to the single shot at the top. The number of balls in a complete triangular pile is evidently equal to the sum of the series 1, 1+2, 1+ 2+ 3, 1+2+3 +4, &c. to 1+ 2+

on one side of the base.

+n, n denoting the number of balls

But from example 3d, last article, we find the sum of n terms of the series,

vc., n2, which we see, from example 4th of the last article, is

(m+1)

213. The complete oblong pile has (m + 1) balls in the upper layer, 2. (m + 2) in the next layer, 3 (m+3). in the third, and so on: hence, the number of balls in the complete pile, is given by the formula deduced in example 5th of the preceding article,

214. If any of these piles is incomplete, compute the num ber of balls that it would contain if complete, and the number that would be required to complete it; the excess of the former over the latter, will be the number of balls in the pile. The formulas (1), (2) and (3) may be written,

1 n (n + 1)

2

angular face of each pile, and the next factor, the number of balls in the longest line of the base, plus the number in the side of the base opposite, plus the parallel top row, we have the following