Some series which cannot be treated by any of the above methods will be discussed in Chapter (xvii.). 3. Sum the series 4+ 16 + 40+ 88+, &c., to r terms; also a + (a + b) + (a + b + br) + (a + b + br + br2) +, &c., to n terms, and ad inf. when possible. 13+23+33 +, &c. &c. + n3; a3 + (a + b)3 + (a + 2b)3 + to n terms; 2.4.6 + 3.5.7 +, 4. Find the sum of each of the following to n terms :— &c. (n − 3) (n − 2). 6. Also the series whose nth term = 7. Find the nth term, and the sum of n terms of a series in which each term is the square of the preceding one, the first term being given. 8. Find the sum of the following, each to n terms : 2.32.73.42.8 +4.52.9 +, &c. ; 4.72.99.122.14 + 14.172.19+, &c.; 23 — 33 + 43 — 53 + 63 — 73 +, &c. What does the last series become when n is indefinitely great? 1.32 +2.42 +3.52+, &c. 9. Find the sum of 200 terms of the series 6+13 +22 + 33 + 46 +, &c. Also of n terms of the two following:- 5+6+8+11 + 15 +, &c., and 7+ 11+18 + 28 + 41 +, &c. 10. Find the 500th term of the series 9 + 28 + 53 +84 +, &c., and the nth term of 2 - 11 - 23 - 34, &c. 11. Which term of the series 7+12+19 +28+, &c., amounts to 206? 12. Find the sum of 2n terms of the series whose 7th term is 3r22r8. 13. Find the sum of 3n terms of a series whose first term is 1, and of which each term exceeds the preceding by 2n2 (n being the order of the term). 14. Find the sum of the infinite series a + ar + (a + ab)r2 + (a + ab + ab2)r3 +, &c. ; r being less than 1, and b less than r−1. 15. Pearls in Ceylon are sorted into 10 classes, by being riddled through perforated trays, the number of perforations in each tray being 19, 37, 61, 21, &c. Find the nth term of this series, and the sum of n terms. 16. If the series of odd numbers be arranged as follows, prove that the sum of the numbers in the nth line is n3, and that the sum of the first n lines is In "1.2 Jn (n + 1) 2 CHAPTER XV. PERMUTATIONS AND COMBINATIONS. 161. Def.-The "Permutations" of a given number of things are the different arrangements which can be made of them in respect of their order. Thus abc, acb, bac, bca, cab, cba, are the permutations of the three letters abc taken all together. Again, ab, ba, ac, bc, ca, cb, are the permutations of abc taken two together. 162. Def.-The "Combinations" of a given number of things are the different collections which can be made of them without regard to the order in which they are placed. Thus ab, ac, bc, are all the combinations of abc taken two at a time. To find the number of permutations of n things taken two at a time. Let the n things be the letters abcd, &c.; n in number. Now it is evident that a may be placed before each of the other letters, making the group Also, b may be placed before each of the others, making the group Similarly, all the letters will stand first in their turn, and each group will consist of n- 1 permutations, and the number of groups will be n. Hence the whole number of permutations of n things will be To find the number of permutations of n things, taken three at a time. Take the n 1 letters b, c, d, &c. The number of permutations of these, taken two together, will be 2) by formula (1), and they are represented in the table (n − 1) (n below. &c. 2 terms of these, we then have to n 2 terms to n 2 terms ... to n 2 terms There are thus (n − 1) (n − 2) permutations of three letters together, in which a stands first. Similarly, it may be shown that there are (n − 1) (n − 2) permutations, in which b stands first, and so on for each of the n letters a, b, c, d, &c. Therefore, the whole number of permutations will be We shall now show generally that the number of permutations of n things taken r together is Let us assume that the number of permutations of the n letters a, b, c, d, &c., taken r |