Substituting these values for a, da, da, &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, d. 18, d= 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, d,= 0, d1 = 0, &c., &c.

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

S=

1.2 3

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, d2 = 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,

Second, in the Square Pile.

212. The square pile is formed, as shown in the figure. The number of balls in the top layer is 1; the number in the second layer is denoted by 22; in the next, by 32, and so on. Hence, the number of balls in a pile of n layers, is equal to the sum of the series, 12, 22 32,

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

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 number 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,

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

RULE.

Add to the number of balls in the longest line of the base the number in the parallel side opposite, and also the number in the top parallel row; then multiply this sum by one-third the number in the triangular face; the product will be the number of balls in the pile.

EXAMPLES.

1. How many balls in a triangular pile of 15 courses?

Ans. 680.

2. How many balls in a square pile of 14 courses? and how many will remain after 5 courses are removed?

Ans. 1015 and 960.

3. In an oblong pile, the length and breadth at bottom are respectively 60 and 30: how many balls does it contain ?

Ans. 23405.

4. In an incomplete oblong pile, the length and breadth at bottom are respectively 46 and 20, and the length and breadth at top 35 and 9: how many balls does it contain? Ans. 7190.

5. How many balls in an incomplete triangular pile, the num ber of balls in each side of the lower course being 20, and in each side of the upper, 10?

6. How many balls in an incomplete square pile, the number in each side of the lower course being 15, and in each side of the upper course 6?

7. How many balls in an incomplete oblong pile, the numbers in the lower courses being 92 and 40; and the numbers in the corresponding top courses being 70 and 18?

in which a, b, c, d, &c., are positive whole numbers, is called a continued fraction: hence,

A CONTINUED FRACTION has 1 for its numerator, and for its denominator, a whole number plus a fraction, which has 1 for its numerator and for its denominator a whole number plus a frac tion, and so on.

216. The resolution of equations of the form