PILING OF BALLS AND SHELLS. (192.) Balls and shells are usually piled in three different forms, called triangular, square, or rectangular, according as the figure on which the pile rests is triangular, square, or rectangular. (1.) A triangular pile is formed by continued horizontal courses of balls or shells laid one above another, and these courses or rows are usually equilateral triangles whose sides decrease by unity from the bottom to the top row, which is composed simply of one shot; and hence the series of balls composing a triangular pile is where n denotes the number of courses in the pile. (2.) A square pile is formed by continued horizontal courses of shot laid one above another, and these courses are squares whose sides decrease by unity from the bottom to the top row, which is also composed simply of one shot; and hence the series of balls composing a square pile is 1+4+9+16+25+.... n2, where n denotes the number of courses in the pile. (3.) The rectangular pile may be conceived to be formed from a square pile, by laying successively on one face of the pyramid a series of triangular strata, cach consisting of as many balls as the face itself contains, and the number of these added triangular strata is always one less than the number of shot in the top row; therefore, if n denote the number of courses, and m+1 the number of shot in the top row, the series composing a rectangular pile is (m+1)+2(m+2)+3(m+3)+4(m+4)+ .... n(m+n) (4.) The number of balls in a complete triangular or square pile must evidently depend on the number of courses or rows; and the number of balls in a complete rectangular pile depends on the number of courses, and also on the number of shot in the top row, or the amount of shot in the latter pile depends on the length and breadth of the bottom row; for the number of courses is equal to the number of shot in the breadth of the bottom row of the pile. Therefore, the number of shot in a triangular or square pile is a function of n, and the number of shot in a rectangular pile is a function of n and m. (5.) If the general term of any series of numbers be of the mth degree, the sum of all the terms of such scries will be of the (m+1)th degree; because the general term of any progressively increasing series being a function of n of the mth degree, the sum of such series evidently cannot exceed n times the general term, that is, it cannot exceed n times a function of n of the mth degree, and therefore the function itself must of the (m+1)th degree. EXAMPLES. (1.) Sum 1+2+3+4+5+. .... to n terms. Here n, the general term of the series, is of the first degree, and therefore the function expressing the sum of the series is of the second degree; and hence we assume 1+2+3+4+ n=Pn2+Qn+R. .... Now this equation must be true for every value of n; hence, wher n=1 we have P+ Q+R=1 n=2...... 4P+2Q+R=1+2 ..... • =1 .... (1) (2) (3) n=3 Assume 1+3+6+10+15+... "(n+1)=Pn2+Qu2+Rn+S...... 2 and since there are four coefficients to be determined, we must have a corresponding number of independent equations; hence when n=1 we have P+ Q+ R+S = 1 = 1 (1) = 4 (2) =10 (3) 64P+16Q+4R+S = 1+3+6+10=20 (4) 27P+9Q+3R+S= n=4 Then (2)—(1) gives (5) (6) (7) (8) = 4 (9) 2Q=3-12P = 1 =1 P= R-3-7P-3Q = + S=1-P-Q-R= 0 ... S=0 Hence 1+3+6+10+15+ ... n(n+1)='n3 + 2n2 +2n = 12 (n2+3n+2) •n(n + 1) (n + 2). Formula for a square pile. (3.) Sum n terms of the series 1+4+9+16+25+. . . . n2. Assume 1+4+9+16+25+... n2=Pn3+Qn2+Rn+S; then, as before, P+ Q+ R+S=1 8P+ 4Q+2R+S=1+4 27P+9Q+3R+S=1+4+9 =1 = 5 =14 64P+16Q+4R+S=1+4+9+16=30 and from these four equations we find, by continued subtraction, P='|, Q=1, R=! and S=0; therefore Formula for a rectangular pile. (4.) By Art. 192, we have the number of shot in a rectangular pite m+square pile _n(n+1), = 2 2 3 Let (m+1)+2(m+2)+3(m+3)+...... n(m+n)=Pn3+Qn2+Rn+S. P+Q+ R+S=(m+1) then 8P+ 4Q+2R+S=(m+1)+2(m+2) 27P+9Q+3R+S=(m+1)+2(m+2)+3(m+3) = m+ 1 = 3m+ 5 = 6m+14 64P+16Q+4R+S=(m+1)+2(m+2)+3(m+3)+4(m+4)=10m+30 and from these four equations we find, as before, P='}', Q='(m+1), R=1'(3m+1), and S=0; hence (m+1)+2(m+2)+3(m+3)+...n(m+n)='n3+m+1n2 + 3m+1 Hence we have the subsequent expressions for (S) the number of balls Now n(n+1) is the number of balls in the triangular face of each pile, and 2 the other factor is the sum of the balls in one side of the base, and the two parallel rows at the top and opposite side of the base; hence we have also this general rule, which, as well as the formulæ, should be committed to memory. Rule. Add to the number of balls or shells in one side of the base, the numbers in its two parallels at bottom and top (whether row or ball), and the sum multiplied by one-third of the slant end or face gives the number of balls in the pile. EXAMPLES. (1.) How many balls are in a triangular pile of 15 courses? Ans. 680. (2.) A complete square pile has 14 courses: how many balls are in the pile, and how many remain after the removal of 5 courses? Ans. 1015 and 960. (3.) In an incomplete rectangular pile, the length and breadth at bottom are respectively 46 and 20, and the length and breadth at top are 35 and 9: how many balls does it contain? Ans. 7190. (4.) The number of balls in an incomplete square pile is equal to 6 times the number removed, and the number of courses left is equal to the number of courses taken away: how many balls were in the complete pile ? Ans. 385. (5.) Let h and k denote the length and breadth at top of a rectangular truncated pile, and n the number of balls in each of the slanting edges; then, if B be the number of balls in the truncated pile, prove that B=" { 2n2+3n(h+h)+6hk—3(h+k+n)+1}. SUMMATION OF SERIES. 193. By a process similar to that we have employed in finding the number of shot in a pile, we may find the general term, as well as the sum of various other series; but we proceed to THE DIFFERENTIAL METHOD. Let a, b, c, d, e, . . . . be a series of terms, in which each term is less than the succeeding one; and, taking the successive differences, we have Putting d1, da, da, d, ..... for the first terms of the first, second, third, fourth, . . . . differences, we have d-3c+3b-a = ds. d = a + 3d, + 3d + dz e-4d+6c-4b+a = d1 :. e = a + 4d, + 6d2 + 4d, + di Hence the (n+1)th term of the proposed series is evidently and, therefore, the nth term is (by writing n-1 for n) a+(n-1)d,+ (n−1) (n—2) d2+ 1.2 .... (n−1) (n—2) (n—39) dst .......... (1) 3 be two series, of which the (n+1)th term of the latter is obviously the sum of n terms of the former; but the first terms of the first, second, third, fourth, differences in the latter, are a, b-a=d1, c-2b+a=d2, d—3c+3b—a, =ds, &c.; hence the (n+1)th term of the latter series, or the sum of n terms of the former is, by eq. (1) above, n(n−1)(n−2)d+ n (n−1) (n−2) (n— 1. 2. 3 n (n−1) (n—2) d2+' n(n−1)d, +2 Or s=na+ 1.2 J. 2. 3 1.2. 3 .4 n (n−1 ) (n—2) (n−3), !dit.. 2dst.. 1. 2. 3. 4 EXAMPLES. (1.) To what is 1.2+2.3+3.4+4.5+. (2) Find the sum of n terms of the series 1, 23, 33, 43, 53, &c. (3.) Find the sum of n terms of the scries 1, 4, 10, 20, 35, &c. (4.) To what is 1.2.3+2.3.4+3.4.5+.... n (n+1)(n+2) equal? (5) Sum n terms of the series 1, 3, 5, 7, 9, 11, &c. . . . (6.) Find the sum of 15 terms of the series 1, 4, 8, 13, 19, &c. (7) Sum 8 terms of the series 1, 21, 3a, 4′, 5′, 6a, &c. |