ART. 329. To find the number of balls in a triangular pile. A triangular pile, as V-ABC, is formed of successive horizontal courses of the form of an equilateral triangle, such that the number of A balls in the sides of these courses, decreases continually by unity, from the bottom to the single ball at the top. V ་ལ་་་ B If we commence at the top, the number of balls in the respective courses will be as follows: and so on. Hence, the number of balls in the respective courses is 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5, and so on; or 1, 3 6 10 15 Hence, to find the number of balls in a triangular pile, is to find the sum of the series 1, 3, 6, 10, 15, &c., to as many terms (n) as there are balls in one side of the lowest course. By applying the formula (Art. 327) to finding the sum of n terms of the series 1, 3, 6, 10, &c., we have a=1, D,=2, D=1, and D.=0. Hence, the formula na+n(n−1)D,+n(n−1)(n−2)D2 1.2.3 _n(n—1) (n—2) ×1=n+n2—n+' 2X3 becomes _n3—3n2+2n (A) __n3+3n2+2n__n(n2+3n+2)___n(n+1)(n+2) ART. 330. To find the number of balls in a square pile. A square pile, as V-EFH, is formed of successive square horizontal courses, such that the number of balls in the sides of these courses, decreases continually by unity, from the bottom to the single ball at the E top, V G H If we commence at the top, the number of balls in the respective courses will be as follows: 14. 2nd. 3rd. 4th There and so on. Hence, the number of balls in the respective courses is 12, 22, 32, 42, 52, &c., or 1, 4, 9, 16, 25, and so on. fore, to find the number of balls in a square pile, is to find the sum of the squares of the natural numbers 1, 2, 3, &c., to as many terms (n) as there are balls in one side of the lowest course. But the sum of the series 1, 4, 9, 16, &c. (see example 2, page 288), is ART. 331. To find the number of balls in a rectangular pile. A rectangular pile, as EFDBCA, is E formed of successive rectangu A to the single row of balls at the top. If we commence at the top, the number of balls in the breadth of the first row is 1, of the second 2, of the third 3, and so on. Also, if m+1 denotes the number of balls in the top row, the number in the length of the second row will be m+2, in the third row m+3, and so on. Hence, the number of balls in the respective courses, commencing with the top, will be 1 (m+1), 2(m+2), 3(m+3), and in the nth course n(m+n). Therefore, the number of balls (S) in a complete rectangular pile of n courses will be S=1(m+1)+2(m+2)+3(m+3)+.. ... +n(m+n) m(1+2+3+4...+n)+(12+22+32+42+ ··· +n2); ... but the sum of n terms of the series in the first parenthesis, (Art. 327,) is n(n+1), and the sum of n terms of the series in the second parenthesis has just been found (Art. 330) to be n(n+1)(2n+1) 6 ; hence, by substitution, we have mn(n+1)__n(n+1)(2n+1)__n(n+1)(3m+2n+1) (C). S: 2 6 6 Here m+n represents the number of balls in the length of the lowest course. If we put m+n=l, we have 3m+2n=31—n; substituting this for 3m+2n, in the preceding formula, it becomes S=n(n+1)(31—n+1). 6 It is evident that the number of courses in a triangular or square pile, is equal to the number of balls in one side of the base course, and in the rectangular pile to the number of balls in the breadth of the base course. ART. 332. Collecting together the results of the three preceding articles, we have for the number of balls In formulæ (A) and (B), n denotes the number of courses, or the number of balls in the base course. In formula (C) n denotes the number of balls in the breadth of the base course, and the number in the length. The number of balls in an incomplete pile is evidently found by subtracting the number in the pile which is wanting at the top, from the whole pile considered as complete. EXAMPLES FOR PRACTICE. 1. Find the number of balls in a triangular pile of 15 courses. Here n=15, and substituting this value instead of n in formula A, (Art. 332), we have the number 15(15+1)(15+2)_15X16X17_680. Ans. 2X3 6 2. Find the number of balls in an incomplete triangular pile of 15 courses, having 21 balls in the upper course. Here we must first find the number of shot in one side of the upper course. From the illustrations in Art. 329, it is evident that the number of balls in any triangular course, is equal to the sum of the natural numbers 1, 2, 3, &c., to the number (n) in one side. Now the sum of the numbers 1, 2, 3, &c., to n, is (Art. n(n+1)= 327) n(n+1); ; hence, )=21, or n2+n=42, from which 2 2 (Art. 231) we find n=6, and therefore 5 courses have been removed from the pile; hence, by formula A, (Art. 332), the number of balls in the pile considered as complete, is 20×21×22=1540, and the number in the pile removed is 2×3 5×6×7 2X3 =1505. =35 .. the number in the incomplete pile is 1540—35 3. Find the number of balls in a square pile of 15 courses. Ans. 1240. 4. Find the number of balls in a rectangular pile, the length and breadth of the base containing 52 and 34 balls respectively. Ans. 24395. 13 5. Find the number of balls in an incomplete triangular pile, a side of the base course having 25 balls, and a side of the top 13. Ans. 2561. 6. Find the number of balls in an incomplete triangular pile of 15 courses, having 38 balls in a side of the base. 3. Ans. 7580. 7. Find the number of balls in an incomplete square pile, a side of the base course having 44 balls, and a side of the top 22. Ans. 26059. 8. The number of balls in the base and top courses of a square pile are 1521 and 169 respectively; how many are in the incomplete pile. Ans. 19890. 9. The number of balls in a complete rectangular pile of 20 courses is 6440; how many balls are in its base? Ans. 740. 10. The number of balls in a triangular pile is to the number in a square pile having the same number of balls in the side of the base, as 6 to 11; required the number in each pile. Ans. 816, and 1496. 11. How many balls are in an incomplete rectangular pile of * 115x, 36+7=43.17+7=24 mot allow the longesthrile wide of lower course 8 courses, having 36 balls in the longer side, and 17 in the 3 ART. 333. INTERPOLATION OF SERIES. Each of the various tables employed in the different departments of science, may be regarded as the terms of a mathematical series. These tables are generally calculated from particular formulæ, but in many cases the computations are so very laborious, that only certain terms at regular intervals, are calculated, and the intermediate ones are derived from these by a process termed Interpolation. Also, in many investigations values of the quantities in the tables are required, intermediate between those given, or extending beyond them. These, likewise, are determined by Interpolation. The principle on which Interpolation is founded is that explained in Art. 326; that is, having certain terms of a series given, to find the nth term. To do this with entire accuracy, requires that we should have such a number of terms of the series given, that we can obtain an order of differences equal to zero. In most cases, however, the differences, D,, D2, D3, &c., do not vanish, but become so small that their omission after D2, or D3, causes no sensible error in the result, and we obtain what is termed, approximate values of the required quantities. ART. 334. When the 3rd order of differences of any given series of quantities vanishes, or becomes very small, then (Art. 326) we have the equation -a+3b-3c+d=0, and any of the quantities a, b, c, or d, may be found, when the other three are given. Similarly, if the fourth differences vanish, then a-4b+6c-4d+e=0. Ex. Given /25-2.92401, 3/26-2.96249, 3/27=3, 3/29 3.07231, to find the cube root of 28. Here four quantities are given to find a fifth, therefore, supposing the fourth order of differences to vanish, we have a-4b+6c-4d+e=0, where d is the term to be interpolated; hence, 4d=a+6c+e-46=2.92401+18+3.07231-11.84996 =12.14636, where d, or 3/28=3.03659, which is true to .00001. |