4. Find the sum to n terms, and ad infinitum, of the following 2 2 1 2.3.4 4.5' 3 &c.; 5. Sum the following to n terms, and ad infinitum :— (2a2 — b2) (3a2 – b2) + (3a2 — b2) (4a2 — b2) + (4a2 — b2) (5a2 — b2), &c. 1. Find the number of balls in a triangular heap, the number of balls in one side of the base being 34. 2. Find the number of balls when the number in one side of the base is 50, also when it is 70. 3. Find the number of balls in the following complete triangular heaps: 1o. When the number of layers is 12. 4. Find the number of balls in an imperfect triangular heap, there being 36 balls in one side of the base, and the number of layers being 16. 5. An imperfect triangular heap of balls has 50 balls in each side of the base, and 24 in each side of the top layer. Find the number of balls. 6. In an imperfect triangular heap of 6 layers the top layer contains 76 balls. Find the whole number of balls in the heap. 7. 1540 balls are to be made into a complete triangular heap. Find the number of balls in each side of the base. 8. 920 balls are to be piled in an incomplete triangular heap of 8 layers. Find the number of balls in each side of the base. 9. Two complete triangular heaps of balls contain 1495 balls, and the greater heap contains 1165 balls more than the other. Find the number of balls in the base of each. 10. The base of a triangular heap of balls contains 20 balls more than the next layer. Find the number of balls in the base. 11. How many balls are there in a complete square heap when the number in one side of the base is 48? 12. Find the number of balls in the following square heaps, which have in one side of the base 100, 55, and 23 balls respectively. 13. If a complete square heap consist of 30 layers, how many balls are there in the heap? 14. If the middle layer of a complete square heap contain 121 balls, find how many there are in the whole heap. 15. If the middle layer of a complete triangular heap contain 66 balls, how many rows are there? 16. Find the number of balls in an imperfect square heap of 10 layers, the number in the base being 1600. 17. An imperfect square heap of balls contains 10 layers, and the number of balls in the base is four times as great as the number in the top layer. Find the whole number of balls. 18. How many balls of 6 inches diameter can be piled on a space of ground 10 yards square, the heap not being more than 6 feet high. 19. How many balls are there in the base of a square heap which contains 2870 balls. 20. The number of balls in a complete square heap the number of balls in a complete triangular heap :: 7 : 4. Find the number of balls in each heap. 21. Find the number of balls in a rectangular pile which has 24 balls in the length, and 16 balls in the breadth of the base. 22. Find the number of balls in the two following heaps: 1°, when the base has 105 in the length and 20 in the breadth; 2°, when it has 40 in the length and 30 in the breadth. 23. The top row of a rectangular pile has 20 balls, and there are 10 courses. Find the number of balls in the pile. 24. Find the difference between the number of balls in a square pile of 50 courses, and two complete rectangular piles each having 50 in the length and 25 in the breadth of the base. 25. Find the number of balls in an incomplete rectangular heap, whose base contains 40 × 30 balls, and which has 10 courses. 26. If the base of an incomplete rectangular pile of 10 courses contain 1911 balls, and the upper course contain 1131 balls, find the number of balls in the heap. 27. Compare the number of balls in a complete square pile with the number in a complete triangular pile which has the same number in one side of its base as there are in one side of the square base, and find the numbers in one side of the base of each, of which the ratio is 1.88. 28. If n be the whole number of balls in a complete square pile, show that the number of courses in the pile is the integer next below the cube root of 3n. 29. If the pile be triangular, show that the number of courses is the integer next below on. APPENDIX. APPLICATION OF THE METHOD OF FINDING THE To adapt the method of Art. (43) to numbers, it is necessary to observe that every number may be put in the form of a series of powers of 10. Thus 5184 5.133 +1.102 +8.10 + 4 which is of the form = 51.1028.10 +4 = 49.102 +28.10 + 4 49x2+28x + 4 .. 7.10+2 or 72 is the square root of 5184. It would however be too laborious to begin by putting large numbers into the form of a perfect algebraical square; instead of this, we may proceed as in the subjoined example. To find the square root of 75625. 7.10 +5.103+6.102+2.10+5 (2.102+7.10+5 4.104 From a careful study of the above example, we derive the following RULE for the extraction of the square root of a numerical quantity. (a.) Set a point over the unit's figure, and over every alternate figure to the right and left. (b.) Find, by inspection, the greatest number whose square is not greater than the number expressed by the figures in the first period. (c.) Set this number in the quotient, and its square under the first period. (d.) Subtract, and take down the two figures in the next period for a dividend. (e.) Double the quotient for a new trial divisor. (f.) Cut off the right hand figure of the dividend, and see how many times the trial divisor is contained in the number expressed by the remaining figures of the dividend. (g.) Write the number in the quotient and in the divisor after the former figures. (h.) Multiply the divisor thus formed by the last number in the quotient. (i.) Subtract, take down the two figures of the next period, and repeat the processes (e.), (f.), (g.), (h.), until all the figures are taken down. Examples worked out. 1. Find the square root of 430.5625 430.5625 (20.75 4 40) .30 00 407) 3056 4145) .20725 20725 Note. In extracting the square root of a decimal, or of a quantity made up of a whole number and a decimal, the number of decimal places in the root is the same as the number of periods or points in the decimal part of the original quantity. 2. Find the square root of to four places of decimals. 2. Extract to 5 decimal places the square roots of the following numbers: 1 2 2: .002; 13.5; 2.025; .0000007; 28.9; 5'9 ; 31; .5; APPLICATION OF THE METHOD OF FINDING THE Take as an example the number 12167. Now 12167 = 12.1031.102 +6.10 +7 Treating this as an algebraical quantity, the process of finding the cube root will be as follows: 12.103+1.102 +6.10 +7 (2.10 +3 4.103 +1.102 +6.10 +7. = 41.1026.10 +7 |