1st. The triangular pile is formed thus: A layer of balls is placed horizontally on the ground in such a manner that the centres of the balls which are at the angles form an equilateral triangle. On this layer is arranged a second, each ball of which is placed in the hollow between three balls of the first layer. This second layer has also the form of an equilateral triangle, but its side contains one ball fewer than the side of the first layer. Layer is placed upon layer in this manner, the side of each containing one ball fewer than the side of the layer on which it rests, until, by successive diminutions of one, the side of the last or uppermost layer (and by consequence that layer itself) contains only one ball. The pile thus formed is a pyramid, having an equilateral triangle for its base; and it is evident that since the uppermost layer of a pyramid of this form contains one ball, the side of the second layer must contain two, the side of the third, three, . . . . . . and the side of the nth, n balls. Denoting by N the number of balls contained in one side of the equilateral triangle which forms the base of a triangular pile, and by Y the number of balls in the base, it is evident that Y=1+2+3+ But this is the sum S, of formula 4 (Art. 192); Therefore Y= N2+N +N. If, in this expression, N is successively replaced by the numbers 1, 2, 3,..... the number of balls in the successive layers, beginning from the top of the pile, will be obtained. 12+1 2 These are, in the first, 2 22+2 2=1. in the second, in the third, Whence the sum X of the whole number of balls contained in the pile is 12+12+232 +3 X= + 2 2 + + 2 Which expression may be otherwise arranged thus, 1+2+3+....+N 1+2+3+ The numerator of the first term of the second member of this equation is the sum S (formula 5, Art. 192), and the numerator of the second term the sum S, (formula 4). Therefore The number of balls contained in a triangular pile is found by this formula (7). Example: How many balls are contained in a triangular pile, the side of whose base contains 35 balls? 2d. Quadrangular pile. In the pile of this form the layer of balls which composes the base is arranged in a square; the other layers are also squares. Each layer has one ball fewer in its side than the layer which is immediately below it, and the layers are continued until the last is composed of one ball only, which forms the top of the pile. The quadrangular pile is, therefore, a pyramid with a square base. Denoting by N the number of balls in a side of the base, and by X the number of balls in the pile, it follows from the construction of the pile that the uppermost layer contains 12=1 ball; the second, 2=4, the third, 32=9; ...... the Nth, No. Example: If a side of the base of a quadrangular pile contains 35 balls the whole number of balls in the pile is 3d. Rectangular pile. In this pile the balls in all the layers except the uppermost are arranged in rectangles. Each side of the layer resting on the base contains 1 ball fewer than the corresponding side of the base; and, similarly, the sides of every layer contain each one ball fewer than the corresponding sides of the immediately inferior layer. When the number of layers is equal to the number of balls in the smaller side of the base, a layer is encountered which consists of only a single row of balls. This row forms the top or ridge of the pile. Representing by p+1 the number of balls in the single row, the layer below it (or the second layer) must contain 2 rows of p+2 balls each; the third, 3 rows of p+3 balls each; . the nth, n rows of p+n balls each. ... If, therefore, n be the number of balls in the smaller side of the base, and X the number in the pile, X=1(p+1)+2(p+2)+3(p+3)+ . . . . . +n(p+n); or X=p(1+2+3+.... +n)+(1o +22 +32 + . . . . n2)=pS1+S or 2 6 or, reducing to the same denominator, and observing that 3p and 2n+1 have the common factor n(n+1), In this formula p is the number of balls less one in the single row which forms the ridge of the pile, n is the number in the smaller side, and p+n the number in the greater side of the base. To express X in terms of the number of balls contained in the sides of the base, let p+n=N, and therefore p=N-n, 3p+2n=3N―n; then, by substitution, X_n(n+1) (3N−n+1) 6 9 Example: If the greater side of the base of a rectangular pile contain 50 balls and the less side 35, how many balls are contained in the pile? 4th. Each kind of pile may be truncated, that is, terminated by an upper layer composed of several rows of balls; in this case the pile is the difference of two complete piles. Examples: 1st. How many balls are contained in a triangular pile, the side of whose base contains 10 balls? Ans. 220. 2d. How many balls are contained in two quadrangular piles, the sides of whose bases contain 9 and 12 balls respectively? Ans. First 285, second 650. 3d. If the greater side of the base of a rectangular pile contain 19 balls and the less side 10, how many balls are contained in the pile? Ans. 880. 4th. The side of the base of a truncated triangular pile contains 15 balls SECTION IX. OF CONTINUED FRACTIONS. 194. The name Continued Fraction is given to an expression of the form x=a+1 d+, &c., in which the letters a, b, c, d, . . . express any whole positive numbers, the series of which may be limited or not. An expression whose numerators and denominators are any quantities whatever may have the form of a continued fraction; but continued fractions of which the numerators are 1, and the denominators whole positive numbers, are those whose properties are most deserving of attention. The numbers a, b, c... are termed incomplete quotients, or, simply, quotients; the complete quotients are obtained by taking each incomplete quotient together with the whole of the quantity which is joined to it by the sign+; such is b + 1 c+, &c. are called partial or integrant fractions. 195. If in a continued fraction the first partial fraction, the first and second partial fractions, the first, second, and third partial fractions, &c. are successively taken, different expressions are obtained which are reducible to the form of vulgar fractions; these fractions are distinguished by the name of Converging Fractions or Convergents. In the preceding expression of a continued fraction the first term or partial fraction is a or 1, and the first convergent is the expression a יך a 1 composed of the first and second terms or partial fractions is a+, which reduced to the form of a vulgar fraction is ab+1; ab+1 is the second convergent. b b The expression composed of the first, second, and third terms, or partial fractions, is a+ 1 b+ The 5th, 6th, .... mth convergents may be formed in the same manner; but the process becomes very laborious when the number of partial fractions is considerable. Any convergent can be found by this method of reduction, without the help of the preceding convergents. If the convergents are formed consecutively in the order second, third, fourth, &c., the second being found by the rule for the reduction of a mixed quantity to an improper fraction, the third can be obtained by changing b into b+ in both terms of the second convergent, and making the necessary ab+1 reductions; thus the second convergent is ; changing into b+ In the same manner the fourth convergent can be deduced from the third by changing c into c+ 1 e 1 the fifth from the fourth by changing d into d+, &c. To obtain any convergent by this process it is necessary to form all the preceding convergents. If the first, second, and third convergents are compared it is found that the third can be formed from the second and first by multiplying ab+1 and b, the terms of the second by c, the third quotient, and adding a and 1, the terms of the first, to the products; and if the second, third, and fourth convergents are compared it will appear that the fourth can be formed from the third and second convergents and the fourth quotient in the same manner. To determine whether every convergent of a series can be formed in this manner from the two preceding convergents, and the quotient of the same N N' N" rank with itself, let DDD" be three consecutive convergents, m the quo N" tient of the same rank as the convergent D", and the partial fraction which n N'" Making this substitution, and denoting the new convergent by D N""__N'(m+)+N_(N'm+N)n+N'_N"n+D' D'""' ̄ ̄D' (m + ;)+D ̄(D'm+D)n+D' ̄D′′n+D" N" is formed by this law from the convergents Whence, if the convergent D N N' and the quotient m, it follows that the convergent and the quotient n, by the same law. from the convergents D" D" Now it has been shown that the third convergent is formed from the first and second, and the fourth from the second and third, by it; therefore the law is general. Whence the (n-1)th and nth convergents of a series and the n+1 quotient being given to find the (n+1)th convergent, multiply the terms of the nth convergent by the (n+1)th quotient, and to the products add the corresponding terms of the (n-1)th convergent. In applying this rule, it is convenient to arrange all the quotients a, b, c, ... in a horizontal line, and to place under them the corresponding convergents as they are formed. If, for example, the continued fraction is 1 x=3+ the arrangement of the quotients and convergents is made as follows: When the continued fraction is not terminated (that is, when it is prolonged indefinitely) it follows, from the law of formation of the convergents, that the numerators and denominators form two series which increase to infinity. 196. The convergents are alternately less and greater than x, the value of the continued fraction. For the first convergent is equal to a; and as the fractional part, which is added to a, is neglected, a<x. 1 The second convergent is equal to a+; and since the divisor of 1 is b, a 1 C' quantity less than b+, the divisor in the continued fraction, it follows that the second convergent is greater than x. 1 The third convergent is equal to a++, in which the divisor ¿+ is 1 1 too great; therefore the fraction is too small, therefore a+b+! b+ The same reasoning may be continued indefinitely. <x. 197. The difference between any two consecutive convergents is equal to 1, divided by the product of the denominators of the two convergents. Let N N N N". the same rank as the convergent ; then N"=N'm+N; D′′=D'm+D. |